% This started out as the notes used in 2005 %\documentclass[a4paper,openany]{book} % eps file does not appear % in draft!!! \documentclass[% version=last,% a5paper,% open=any,% cleardoublepage=empty,% 10pt,% %headings=small,% index=totoc,% bibliography=totoc,% reqno,% draft=false,% %draft,% twoside,% BCOR=2mm,% DIV=18,% headinclude=true,% pagesize]%{scrreprt} {scrbook} \usepackage{scrpage2} \pagestyle{scrheadings} \ohead{\pagemark} \ihead{\headmark} \ofoot{} \renewcommand{\captionformat}{\ \ } \usepackage{pstricks,pst-plot} \usepackage{hfoldsty} \usepackage{url} \usepackage{cclicenses} \usepackage[polutonikogreek,english]{babel} %\usepackage[oxonia]{psgreek} \usepackage{amsmath,amssymb} \usepackage{graphicx} % For the photos \usepackage[all]{xy} %\usepackage[notref,notcite]{showkeys} % For use when writing \usepackage{longtable} \usepackage{multicol} \usepackage[mathscr]{euscript} \usepackage{amscd} % commutative diagram % For index \usepackage{makeidx} %\usepackage{showidx} % For use when editing \makeindex %\newcommand{\itech}[1]{\textsl{#1}\index{#1|textsl}} %\newcommand{\idefn}[1]{\textbf{#1}\index{#1|textbf}} \newcommand{\tindex}[1]{\index{#1|textsl}} \newcommand{\dindex}[1]{\index{#1|textbf}} \newcommand{\tindexsub}[2]{\index{#1!#2|textsl}} \newcommand{\dindexsub}[2]{\index{#1!#2|textbf}} % For symbol table %\makeglossary % This command must be removed when the glossary is to % be printed (during the editing stage) %\newcommand{\glossaryentry}[2]{#1\quad #2\\ } % This is used in notes.glo \newcommand{\glossaryentry}[2]{#1& #2\\ } % This is used in notes.glo % In the tex file, the \glossary command, with one argument, puts that % argument as the first argument of \glossaryentry (the second being a % page number) in the notes.glo file. % At the end I took this file and % made from it, by hand, the symbols.tex file. \usepackage{upgreek} \newcommand{\vnn}{\upomega} \usepackage{bm} \newcommand{\sv}[1]{\bm{#1}} % syntactical variable \newcommand{\liff}{\Leftrightarrow} \newcommand{\lto}{\Rightarrow} \newcommand{\Enot}{\text{\sf not-}} \usepackage{stmaryrd} \renewcommand{\land}{\mathbin{\binampersand}} \newcommand{\amp}{\;{}\mathbin{\&}{}\;} \newcommand{\eor}{\nLeftrightarrow} \renewcommand{\phi}{\varphi} \newcommand{\Ey}{\mathcal A} \newcommand{\Np}{\N^+} \newcommand{\Qp}{\Q^+} \newcommand{\Qpp}{\Q^{++}} \newcommand{\Rp}{\R^+} \newcommand{\olt}{\olessthan} \newcommand{\pred}[1]{\operatorname{pred}(#1)} \newcommand{\class}[1]{\mathbf{#1}} % class \newcommand{\Russell}{\class R} % {x:x\notin x} \newcommand{\on}{\class{ON}} % the class of ordinals \newcommand{\cn}{\class{CN}} % the class of cardinals \newcommand{\shstroke}{\mathrel{|}} % Sheffer stroke \newcommand{\psys}[1]{\mathcal{#1}} % proof-system \newcommand{\proves}[1][]{\vdash_{\psys{#1}}} %\newcommand{\nproves}[1][]{\nvdash_{\psys{#1}}} \newcommand{\fcom}{,\;} % comma in list of formulas \let\oldepsilon\epsilon \renewcommand{\epsilon}{\varepsilon} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\inv}{^{-1}} % mult. inverse \newcommand{\divides}{\mathrel{|}} \newcommand{\ndivides}{\mathrel{\nmid}} \newcommand{\radix}{\sqrt{\rule{0ex}{1.5ex}}} \renewcommand{\leq}{\leqslant} \renewcommand{\geq}{\geqslant} % Standard structures \newcommand{\stnd}[1]{\mathbb{#1}} \newcommand{\N}{\stnd{N}} % natural numbers \newcommand{\Z}{\stnd{Z}} % integers \newcommand{\Q}{\stnd{Q}} % rationals \newcommand{\R}{\stnd{R}} % reals \newcommand{\C}{\stnd{C}} % complex numbers \newcommand{\B}{\stnd{B}} % the 2-element Boolean algebra % Symbolic logic \DeclareMathOperator{\arity}{arity} % \newcommand{\qsep}{\;} % follows a quantified variable \newcommand{\Forall}[1]{\forall{#1}\qsep } \newcommand{\Exists}[1]{\exists{#1}\qsep } \newcommand{\existsunique}{\exists!} \newcommand{\Existsunique}[1]{\exists!\,{#1}\qsep } \newcommand{\Frall}[2]{(\forall#1\in#2)\qsep } \newcommand{\Exsts}[2]{(\exists#1\in#2)\qsep } \newcommand{\interpretation}{\mathscr I}% interpretation (part of a %structure) \newcommand{\fv}[1]{\operatorname{fv}(#1)} % set of free variables in #1 \newcommand{\Cn}[2][\lang]{\operatorname{Con}_{#1}(#2)} % logical consequences \newcommand{\conseq}[1]{\operatorname{Con}(#1)} % logical consequences \renewcommand{\models}{\vDash} % for parallelism with the following: \newcommand{\nmodels}{\nvDash} \newcommand{\Fm}[2]{\operatorname{Fm}^{#1}(#2)} % #1-ary formulas \newcommand{\str}[1]{\mathfrak{#1}} % structure %%%% SET THEORY \newcommand{\universe}{\mathcal U} % set-theoretic operators: % nullary: \renewcommand{\emptyset}{\varnothing} % [I like the other symbol better] % unary: \newcommand{\comp}{^{\mathrm{c}}} % set-theoretic complement \newcommand{\pow}[1]{\mathscr{P}(#1)} % power set % binary: \renewcommand{\setminus}{\smallsetminus} % I don't like the standard symbol \newcommand{\symdiff}{\vartriangle} % symmetric difference % [infinitary]: \newcommand{\family}[1]{\mathcal{#1}} % family (of sets) % set-theoretic relations: \newcommand{\included}{\subseteq} % [the name suggests the meaning here] \newcommand{\nincluded}{\not\subseteq} % not included \newcommand{\pincluded}{\subset} % proper inclusion % cardinality \newcommand{\injects}{\preccurlyeq} % injects in \newcommand{\pinjects}{\prec} \newcommand{\equip}{\approx} % equipollent (is in bijection with) \newcommand{\nequip}{\not\approx} % not equipollent \newcommand{\vscr}[1]{#1'} % successor ordinal (v for von Neumann) \newcommand{\size}[1]{\lvert#1\rvert} % cardinality % Model theory \newcommand{\lang}{\mathcal{L}} % a language or signature % relations and functions \newcommand{\conv}[1]{\breve{#1}} % converse of a relation \newcommand{\To}{\longrightarrow} \DeclareMathOperator{\dom}{dom} % domain of a function \newcommand{\setim}[1][]{''{#1}} % image of set under preceding function \newcommand{\setimb}[1]{[#1]} % image of set under preceding function \newcommand{\rest}[1]{\restriction{#1}}% restriction of function to #1 \newcommand{\id}{\mathsf{id}} % identity-map \newcommand{\mapset}[2]{{}^{#1}#2} % set of functions from #1 to #2 \newcommand{\chf}[1]{\chi_{#1}} % characteristic function \newcommand{\coordproj}[2]{\pi^{#1}_{#2}} % coordinate projection, % omitting entry {#2} from an {#1}-tuple \newcommand{\modsim}{/\mathord{\sim}} % modulo the eq-ren \sim % Number theory \newcommand{\scr}[1]{#1^{+}} % successor \DeclareMathOperator{\lcm}{lcm} \newcommand{\parity}[1]{\operatorname{p}(#1)} \newcommand{\rem}[2]{\operatorname{rem}(#1,#2)} % remainder on % dividing #1 by #2 \usepackage{verbatim} % for the comment environment %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\input{../../../../../TeX/format} \usepackage[neverdecrease]{paralist} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amsthm} \newtheorem{theorem}{Theorem}[section] \newtheorem{axiom}[theorem]{Axiom} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{porism}[theorem]{Porism} \theoremstyle{definition} \newcommand{\raisebullet}{\mbox{}\\[-2.3\baselineskip]\mbox{}} % for % use with the following: \newtheorem{example}[theorem]{Example} \newtheorem*{example*}{Example} \newtheorem{examples}[theorem]{Examples} \theoremstyle{remark} \newenvironment{exercise}% {\subsection*{Exercise}}% {} \newenvironment{exercises}[1][]% {\subsection*{Exercises}{#1}\begin{enumerate}}% {\end{enumerate}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\full}{full} % as in full truth-table; it used to be % `proper' \newcommand{\tuple}[1]{\vec{#1}\,} \newcommand{\named}[1]{\widehat{#1}} % truth-value of #1 % For the numbers chapter: \newcommand{\axz}{Axiom~Z} \newcommand{\axu}{Axiom~U} \newcommand{\axi}{Axiom~I} \newcounter{fnexample} \newcommand{\nex}{\stepcounter{fnexample}\thefnexample} \newcommand{\Tur}[1]{\textsf{#1}} % for words in Turkish %\newcommand{\Eng}[1]{\uline{\textsf{#1}}} %for words in English \newcommand{\Eng}[1]{\textsf{#1}} %for words in English %\newcommand{\Gk}[1]{\begin{greektext}#1\end{greektext}} \newcommand{\Gk}[1]{\selectlanguage{polutonikogreek}#1\selectlanguage{english}} \newcommand{\Lat}[1]{\textsc{#1}} \newcommand{\letter}[1]{\textsf{#1}} %for letters as such \newcommand{\annot}[1]{\footnote{#1}} \begin{document} \title{%Introductory Notes\\ on the\\ Foundations\\ of\\ Mathematical Practice} \author{David Pierce} \date{\today} \uppertitleback{\center{ This work is licensed under the\\ Creative Commons Attribution--Noncommercial--Share-Alike License.\\ To view a copy of this license, visit\\ \url{http://creativecommons.org/licenses/by-nc-sa/3.0/}\\ \mbox{}\\ \cc \ccby David Pierce \ccnc \ccsa\\ \mbox{}\\ Mathematics Department\\ Middle East Technical University\\ Ankara 06531 Turkey\\ \url{http://metu.edu.tr/~dpierce/}\\ \url{dpierce@metu.edu.tr} } } \maketitle \tableofcontents \listoffigures %\input{preface} %\setcounter{chapter}{-1} \addchap{Preface} %[2006.01.16: These notes are being edited, following their use in %the first semester of 2005/6.] This book concerns the foundations\index{foundation} of mathematics in two ways: \begin{compactenum} \item this book is about concepts and techniques that all mathematicians use, implicitly or explicitly; \item this book (or parts of it) is intended for use in a first university-level mathematics course. \end{compactenum} More precisely, these notes are originally written for a course called Fundamentals of Mathematics, given at Middle East Technical University in Ankara under the designation Math 111.\footnote{The catalogue description of Math 111 is: \begin{quote} Symbolic logic. Set theory. Cartesian product. Relations. Functions. Injective, surjective and bijective functions. Composition of functions. Equipotent sets. Countability of sets. More about relations: equivalence relations, equivalence classes and partitions. Quotient sets. Order relations: Partial order, Total order, Well ordering. Mathematical induction and recursive definitions of functions. \end{quote}} The notes also offer additional reading for those interested in the topics they discuss. In particular, the notes may be useful for Math 320 (Set Theory) and Math 406 (Introduction to Mathematical Logic and Model Theory) at METU. What are foundations? A wooden house may by built on a stone foundation. A mason lays down the stones; then a carpenter erects the house on top. The carpenter cannot construct the walls and floors of the house before the stone-mason creates a place to set those floors and walls; but the stone-mason cannot create this foundation without knowing what the carpenter intends to place there. So it is with the foundations of mathematics. You cannot do mathematics without a place to start; but you cannot create the starting-point without knowing the mathematics that will proceed from it. This is a paradox---a seeming contradiction. It is not a \emph{real}\index{real} contradiction; but it does suggest that the nascent mathematician (the first-year student) cannot read this book page after page as if it constituted an easy novel. The book might considered as a difficult novel with lots of interrelated events. (However, not every novel has an index or a list of symbols like this one.) Not every section of the book should be studied in sequence during the reader's first encounter. Even if an earlier section \emph{is} required for a later section, still, that earlier section may not be fully comprehensible without some knowledge of the later section. What can the reader do? Read slowly, but jump ahead; reread what you have already read; \emph{think} the whole time, but do not think too much without really knowing what you are thinking about. Talk to classmates; talk to teachers. Read with a pencil. Summarize passages in your own words. Invent your own symbolism (while remembering that communicating with others requires a common symbolism). Read other books on the same subjects. Also: {do exercises.} Create your own exercises. Most sections of the book end with exercises. The student who is in a hurry will find out from a teacher which exercises to work on and will then try to do them immediately, looking back into the sections as necessary for examples. A difficulty in this approach is that most exercises here do not have unique correct \emph{answers;} they have \emph{solutions,} some of which are better than others. Finding the best solutions---even acceptable solutions---will require reading, thinking, and experience. Still, many of the exercises can be approached as puzzles\index{puzzle}: they do not need deep insight into the nature of things, but aim only to develop facility with some basic ideas. Most exercises here could not very well be cast as multiple-choice questions. In a multiple-choice question, if you can somehow figure out the correct answer, even without being able to say how you did it, your answer is still $100\%$ correct. Here, correct solutions to problems will carry \emph{within themselves} the reasons why they are correct. There are no answers at the back of the book. Problems here can have more than one correct solution; \emph{you} should be able to tell whether a particular solution is correct. It is true that you may fail to notice some mistakes; the only way to avoid this is \emph{experience,} not desire or {will.} Somebody who does not know a language very well will not avoid mistakes just by trying hard: s/he\footnote{The construction \Eng{s/he} can be pronounced as \Eng{she or he} (or as \Eng{he or she}). English has not evolved a generally accepted singular pronoun that refers to humans of either sex: it lacks the \Tur{o(n)} of Turkish. In the fourteenth century, according to the Oxford English Dictionary (OED) \cite{OED}, the second-person plural pronoun \Eng{you} began to be used respectfully in place of the singular pronoun \Eng{thou}, just as the Turkish \Tur{siz} replaces \Tur{sen}. In the same way, currently, some people use \Eng{they} with a singular sense. Other people are bothered by this usage, and they may insist that \Eng{he} can refer to humans of unknown sex. The original OED does not recognize this usage. However, the OED does claim that \Eng{she} comes from a different base from \Eng{he}, because the feminine form derived from the base of \Eng{he} was too much like the masculine form.} must \emph{practice.} Likewise with games: even if you memorize all of the moves of chess and think real hard, you will not play a good chess-game at first. Depending on how seriously you take mathematics, you can see this book as lessons in a language or a game. It would be worthwhile for the reader to have a look at Euclid's \emph{Elements}. (Heath's English translation from the Greek is \cite{MR17:814b}---see the bibliography at the end of the book. This translation is available in print and in various places around the Web.) The present book does not share much \emph{content} with the \emph{Elements;} but Euclid's work does establish a sort of foundation or prototype for the mathematics of his and all succeeding generations, including our own. Euclid wrote the \emph{Elements}, the original textbook of mathematics, some 2300 years ago.\footnote{Euclid practiced mathematics in Alexandria around 300 \textsc{bce}, probably having learned mathematics in Athens from the students of Plato \cite[vol.~I, pp.~1~f.]{MR17:814b}.} This textbook is still in use in some classrooms today. It consists of 13 books. You are not likely to read all of them; as with the present work, you will jump around, reading what you are interested in, perhaps with the guidance of a teacher. Indeed, perhaps Euclid expected few people to read his work unaided. His work does bring the reader instantly into real mathematics; but it also sets a standard for \emph{spareness} (terseness, economy) of mathematical composition. The \emph{Elements} contains no commentary, no guidance for the reader. After a few definitions and \textsl{axioms,}\tindex{axiom} the work consists solely of \textsl{propositions}\tindex{proposition} and their \textsl{proofs.}\tindex{proof} Euclid does not \emph{tell} you, but he \emph{shows} you what proofs of propositions are. The present book contains more than just propositions and their proofs; but it \emph{does} contain these. Each proof here is labelled as such, and it ends with a little \label{page:box} box. (The first example is on p.~\pageref{first-proof}.) The propositions and proofs in the book consist of sentences of ordinary language, with some abbreviative symbolism (as well as the symbolism required by what the proofs are \emph{about}). Such proofs might be called \emph{informal,} because ordinary language is itself informal. Grammatical rules for English or Turkish or any other human language can indeed be formulated, and the conscientious speaker or writer will try to follow them; but it seems impossible to formulate grammatical rules that are obeyed by, and only by, everything that one wants to say. Informal proofs are to be distinguished from \textsl{formal proofs.}% \tindexsub{formal}{--- proof}% \tindexsub{proof}{formal ---} Again, the notion of proof itself---\emph{informal} proof---is over two thousand years old; but the notion of a \emph{formal} proof dates only from the 1920s.\footnote{Perhaps the invention can be attributed to Hilbert \cite[\S07, n.~110]{MR18:631a}.} This book \emph{tells} you, as well as shows you, what formal proofs are. Briefly described, a formal proof is a list of sentences of an \emph{artificial} language;\index{language!artificial ---}\index{artificial language} but such a list must satisfy certain requirements. The last sentence on the list is what the formal proof is a proof \emph{of}: it is what the proof \emph{proves}. A machine could check whether a given list of sentences is a formal proof. To establish the truth of an interesting proposition, a formal proof is practically never called for. However, if it is held to the highest standard, an informal proof of some proposition $P$ can be seen as an argument that a formal proof of $P$ could in principle be written. It will be an exercise in this book to write some formal proofs; but the ultimate goal is the ability to check the validity of \emph{informal} proofs (like Euclid's, or any later mathematician's), \emph{and} the ability to write one's own (informal) proofs. I assume that you, the reader, have some experience with high-school algebra, and specifically with the algebra of the \textsl{integer}s\tindexsub{number}{integer}. Then you can prove an identity\index{identity} like \begin{equation*}\tag{$0.1$} x^3+y^3=(x+y)(x^2-xy+y^2) \end{equation*} (by multiplying out the right member and combining like terms). The algebra of the integers serves as a pattern for \textsl{Boolean algebra,} which I shall introduce as the algebra of the numbers~$0$ and~$1$ alone. If one considers these numbers to represent \emph{falsity} and \emph{truth,} then Boolean algebra determines an algebra of \textsl{propositions,} \tindex{proposition} or a propositional \textsl{logic}% \tindexsub{logic}{propositional ---}% \tindexsub{proposition}{---al logic}. After we have propositional logic, we can say something about \textsl{predicate} logic.% \tindexsub{logic}{predicate ---}% \tindexsub{predicate}{--- logic} This logic provides for the analysis of propositions into parts, some of which are \emph{not} propositions. (Some parts of propositions will be \textsl{predicates:}\tindex{predicate} hence the name of the logic.) We cannot define everything precisely until we have the notion of a \textsl{relation.}% \tindex{relation} Relations are certain \textsl{sets;}\tindex{set} they are \textsl{subsets}\tindexsub{sub}{---set}\tindexsub{set}{sub---} of \textsl{Cartesian products}\tindex{Cartesian product}\tindexsub{product}{Cartesian ---} of sets. So all of these things will need to be discussed. A \textsl{function}% \tindex{function} can be defined as a kind of relation. Functions give us a way to say when two sets `have the same size', or are \textsl{equipollent}% \tindex{equipollent, equipotent} (or \textsl{equipotent}). The set of integers has the same size as the set of \emph{even} integers; both sets are \textsl{countably infinite;}% \tindexsub{infinite}{countably ---}% \tindexsub{count}{---ably infinite} but there are strictly \emph{larger}, \textsl{uncountable}% \tindex{uncountable}% \tindexsub{infinite}{uncountable} sets, such as the set of \textsl{real numbers.}% \tindexsub{real}{--- number}% \tindexsub{number}{real ---} The predicate logic given here is more precisely called \textsl{first-order} predicate logic.\tindexsub{logic}{first-order ---}\tindexsub{first}{---{}-order logic} Functions also allow us to give an account of \textsl{first-order logic} in general. The integers have an \textsl{ordering.}\tindexsub{order}{---ing} This is a kind of relation. There is a generalization called a \textsl{partial ordering}\tindexsub{partial}{--- ordering}\tindexsub{order}{partial ---ing}. We shall prove a \textsl{representation theorem}\tindexsub{represent}{---ation theorem}\tindexsub{theorem}{representation ---}, namely the proposition that every partial ordering behaves like the subset-relation (in a clearly defined way). Equality is also a relation and is the motivating example of an \textsl{equivalence-relation}. \tindexsub{equivalence}{---{}-relation} \tindexsub{relation}{equivalence-{}---} The standard sorts of numbers---integers, rational numbers, real numbers, complex numbers---can be formally defined in terms of equivalence-rela\-tions, once one has the \textsl{natural numbers}% \tindexsub{number}{natural ---}% \tindex{natural number} $0$, $1$, $2$, $3$, \dots. The idea of this book is that we do not \emph{really} have these numbers, mathematically, until we can give a logical account of them. This book ends with such an account. The topics of this book are so interrelated that, in any discussion of them, it is hard to avoid the appearance of circularity. This circularity is a part of the foundational aspect of the book. As I say, I assume that the reader is familiar with the integers; but I also say that we do not officially \emph{have} the integers until the end of the book. Yet my supposedly rigorous account of the integers depends on all of the machinery that the book develops first, with the aid of a familiarity with\dots the integers. Our path will have been, not circular, but spiral or rather \emph{helical}, as if along a winding staircase. We start from the integers, and then we return to them, but at a higher (or deeper) level than where we started. \section*{Typography}\label{sect:typography} These printed words are assembled by means of the collection of typesetting programs and packages known as \AmS-\LaTeX. Here, \TeX\ is in Greek letters\footnote{See \cite[p.~1]{Knuth}.}; the same three letters will appear below, in \S~\ref{sect:logic}, in the full Greek name of logic. In the Latin alphabet, the letters are written \Lat{tech}, as in \Eng{technical}. The \AmS\ is the American Mathematical Society. The original \TeX\ program was expanded into \LaTeX\ and independently into \AmS-\TeX; then the benefits of both expansions were combined into \AmS-\LaTeX. The original \TeX\ program distinguishes between ordinary text and mathematical text. In ordinary text, in this book, words are \emph{italicized} for the usual sorts of reasons: they (or their meanings) are being emphasized, they are titles, they are not in the language of the surrounding text, and so forth. I am also making some further distinctions. Technical terms are in \textbf{bold-face} when they are being defined, explicitly or implicitly. Technical terms might simply be \textsl{slanted} if their precise definitions will come later or are simply not needed. Words that are being \emph{talked about} or \emph{mentioned} (and not simply being \emph{used})\footnote{The distinction between the \textsl{use}% \tindex{use} and the \textsl{mention}% \tindex{mention} of a word (or symbol) is attributed to Quine in \cite[\S~08, p.~61]{MR18:631a}. The sentence `A woman or a man is a human' uses the word \Eng{woman}. The sentence `The English word for \Tur{kad\i n} also has five letters' mentions the word \Eng{woman} without using it. The sentence `\Eng{Woman} has five letters' uses the word \Eng{woman} to mention the same word. Such a use can be called \textbf{autonymous,} following Carnap: again, the attribution is in \cite[\S~08, p.~61]{MR18:631a}, where it is said that Frege introduced the practice of indicating autonymous use of words by quotation-marks (inverted commas). By this practice, the last quoted sentence would be ``\,`Woman' has five letters.''} are in \Eng{sanserif}. However, I may not have always been consistent in making these distinctions. Footnotes here are intended to contain only material that is not essential to the main point. They may contain historical information that I have happened to discover, although much of the history of what I am discussing is still unknown to me. %Footnotes may also contain forward references. Labelled proofs here end with boxes, as noted above; labelled examples end with bullets (the first is on p.~\pageref{first-example}). \section*{Acknowledgments} The contents of this book have appeared in several editions: \begin{asparaenum}[1.] \item Some of the material on logic and numbers was first prepared by me in 2001; at the same time, Andreas Tiefenbach prepared notes on sets and relations. Andreas, Belgin Korkmaz, and I taught Math 111 from those notes. \item Andreas and I revised our respective notes, with Belgin's advice, the following year. \item In 2004, in preparing to teach Math 111 that fall along with Ay\c se Berkman and Mahmut Kuzucuo\u glu, I composed my own complete set of notes, drawing on Andreas's notes in giving my own account of sets and relations. \item After that semester, I revised the notes, keeping in mind the experience of Ay\c se, Mahmut and myself, along with impressions from students. Advice from my friend Steve Thomas was also useful for this revision. The notes were used next year, in the fall semester of 2005/6, when I taught Math 111 with Belgin Korkmaz and Turgut \"Onder. \item This new revision is based on that experience. However, there have been many changes, and this book must still be considered as a rough draft, a work in progress. \end{asparaenum} Many of the topics dealt with in this book are also covered by basic texts like \cite{Epp} or \cite{01461249}. I am not trying to write such a book as these are, but I find it useful to look at them. The preface of \cite{01461249} is particularly reassuring, as it describes the many changes that the authors have made in each new edition of their book. My own notes on logic draw from various sources, especially \cite{MR18:631a} and \cite{Burris}; Ali Nesin's book~\cite{Nesin} is an account in Turkish of some of the same material. Set-theory on the level of my coverage seems generally to be found only in more advanced texts like \cite{MR0349389} or \cite{MR1924429}. I use these books, but try to give more elementary exercises than they do. For my notes on natural numbers, \cite{MR12:397m} is inspirational. As a student, I appreciated the style of Spivak \cite{0458.26001}: not condescending, but treating the reader as a fellow mathematician. \chapter*{} \begin{quote} \begin{center} \textsc{Open Your Own Treasure House} \end{center} Daiju visited the master Baso in China. Baso asked: ``What do you seek?'' ``Enlightenment,'' replied Daiju. ``You have your own treasure house. Why do you search outside?'' Baso asked. Daiju inquired: ``Where is my treasure house?'' Baso answered: ``What you are asking is your treasure house.'' Daiju was enlightened! Ever after he urged his friends: ``Open your own treasure house and use those treasures.'' \begin{flushright} Paul Reps and Nyogen Senzaki\\ `101 Zen Stories'\\ \emph{Zen Flesh, Zen Bones}\\ \cite[p.~55]{ZFZB} \end{flushright} \end{quote} %\input{chapter-preliminaries.tex} \chapter{Introduction}\label{ch:prelim} %\renewcommand{\dictumwidth}{0.667\textwidth} %\dictum{ %} \setcounter{section}{-1} \section{Logic}\label{sect:logic} The name of \textbf{logic}\dindex{logic} comes ultimately from the (ancient) Greek adjective \Gk{logik'h}, which is short for \Gk{}) and the three tonal accents %(\Gk{'{\ }}, \Gk{~}, \Gk{`{\ }}) (\Gk{'a}, \Gk{~a}, \Gk{`a}) can be ignored.} \label{fig:Greek} \end{figure} Section~\ref{sect:language} of the book makes a preliminary approach to the notion of a proposition, introducing the terminology of \textsl{axioms}% \tindex{axiom} and \textsl{theorems.}% \tindex{theorem} Section~\ref{sect:sets} introduces the basic terminology of \textsl{sets}% \tindex{set} and \textsl{natural numbers;}% \tindex{natural number}% \tindexsub{number}{natural ---} some of this terminology is used in the review of arithmetic in \S~\ref{algebra}. Arithmetic will be familiar to everybody from school; the main purpose here is to develop a point of view, a way of looking at mathematics, which we shall then apply to logic. Also, the notion of \textsl{arithmetic term}% \tindexsub{arithmetic}{--- term}% \tindexsub{term}{arithmetic ---} introduced in \S~\ref{algebra} will provide an example of a \textsl{proof by induction.}% \tindexsub{proof}{--- by induction}% \tindexsub{induction}{proof by ---} Finally, arithmetic is the setting for some ancient mathematical proofs; these are given as examples in \S~\ref{proofs}. Further investigation into these examples is in \S~\ref{sect:anthyphaeresis}. In \S\S~\ref{sect:parity} and~\ref{sect:connectives}, some operations on the set $\{0,1\}$ are introduced by means of, and by analogy with, the usual arithmetic operations. What these correspond to in ordinary language is discussed in \S~\ref{sect:p-formulas}; further logical analysis of language is in \S~\ref{sect:quantifiers}. The operations on $\{0,1\}$ are essential to the study of mathematical logic as such, which begins in Chapter~\ref{ch:logic}. \subsection*{Exercise} Memorize the Greek alphabet. \section{Language and propositions}\label{sect:language} We are using language right now. We divide up language into \textbf{sentences.}\dindex{sentence} Some sentences, but not all, can be described as \textbf{true}% \dindex{true} or \textbf{false.}% \dindex{false} At least, some sentences are true or false when placed in a \textsl{situation}% \tindex{situation} or \textsl{context.}% \tindex{context} Let us refer to such sentences as \textbf{statements}% \dindex{statement} or \textbf{propositions.}% \dindex{proposition}\footnote{We could make a distinction here: we could let a \textsl{statement} be a bit of language of a certain grammatical form, letting a \textsl{proposition} be the \textsl{meaning}% \tindex{meaning} of a statement. See \cite[p.~26]{MR18:631a}. I am \emph{not} going to try to make such a distinction.} For example, the sentence \begin{center} \Eng{I went to Van last year} \end{center} is a statement (or a proposition). Whether it is true or false depends on who says it and when: the speaker and the time would be the \emph{context} in which the sentence is true or false.\footnote{The context can also include the listener, as when the sentence is \Eng{You went to Van last year.}} We shall mainly be interested in \emph{mathematical} propositions. Such propositions are timeless and personless: their truth or falsity does not change with time or with the person who utters them. Still, in \S~\ref{sect:1st}, we shall see a way in which, strictly, a mathematical proposition must still be placed in a context in order to become true or false. The \emph{belief} that a mathematical proposition is true or false may change with time. Certain mathematical propositions have been accepted as true for many years, only to be found false. For example, Proposition I.16 of Euclid's \emph{Elements} can be called false, even in its context, since its proof relies on unstated assumptions that do \emph{not} follow from the stated assumptions; but this falsehood was not recognized\footnote{See Heath \cite[vol.~1, p.~280]{MR17:814b} for some commentary.} until the nineteenth century. However, the philosopher R. G. Collingwood writes in his autobiography \cite[pp.~31--33]{Collingwood-Auto}: \begin{quote} [Y]ou cannot find out what a man means by simply studying his spoken or written statements, even though he has spoken or written with perfect command of language and perfect truthful intention. In order to find out his meaning you must also know what the question was (a question in his own mind, and presumed by him to be in yours) to which the thing he has said or written was meant as an answer\dots If the meaning of a proposition is relative to the question it answers, its truth must be relative to the same thing. \end{quote} If we \emph{translate} Euclid's work into the kind of formal proofs that will be developed in this book, then indeed we shall find errors or gaps in the proofs. Euclid himself was not writing formal proofs; there was no notion of such a thing for over two thousand years. However, Euclid \emph{was} doing mathematics, and correctly, I would say; but this is for you to judge, \emph{after} reading Euclid himself and understanding his purpose---after understanding the questions he was answering. Euclid's work begins with five propositions that we call \textsl{axioms} or \textsl{postulates}. (He, apparently \cite[p.~442]{MR13:419a}, called them \Gk{a>it'hmata}, that is, requests, demands, or assumptions.) An \textbf{axiom}\dindex{axiom}\dindexsub{proposition}{axiom} is usually a proposition that satisfies two criteria: \begin{compactenum} \item it is \textsl{self-evident;} \item it is useful for proving other propositions. \end{compactenum} In common usage, the first criterion is probably more important; in mathematical usage, the second. A \textbf{self-evident}% \dindexsub{self}{---{}-evident}% \dindexsub{proposition}{self-evident ---} proposition is self-evidently \emph{true:} that is, obviously true without any need of appeal to some other authority. A classical use of the compound word \Eng{self-evident} is found in a certain revolutionary manifesto\footnote{Namely, the Declaration of Independence of the United States of America, written in 1776 by Thomas Jefferson, who, with other signers of the document, possessed other human beings as slaves. In 1945, Vietnamese revolutionaries led by Ho Chi Minh issued a Declaration of Independence that enunciated some of the truths of the American declaration \cite[ch.~18]{Zinn}; this did not prevent a later American invasion.} of the eighteenth century. I transcribe from \cite[p.~15]{Heffner}, with my own formatting: \begin{quote} We hold these truths to be self-evident, \begin{compactenum}[1)] \item that all men are created equal, \item that they are endowed by their Creator with certain unalienable rights, \item that among these are life, liberty and the pursuit of happiness. \item That to secure these rights, governments are instituted among men, deriving their just powers from the consent of the governed. \item That whenever any form of government becomes destructive of these ends, it is the right of the people to alter or abolish it, and to institute new government, laying its foundation\index{foundation} on such principles and organizing its powers in such form, as to them shall seem most likely to effect their safety and happiness. \end{compactenum} \end{quote} Two thousand years earlier, before Euclid even, in the collection of books now known as the \emph{Metaphysics} \cite{Aristotle-XVII}, Aristotle writes of axioms, using the Greek source of our word \Eng{axiom}, namely \Gk{>axi'wma}. This word has the root meaning of \emph{something worthy,} or an \emph{honor.} Aristotle seems to use \Eng{axiom} almost as a synonym of \Eng{principle} (\Gk{>arq'h}) or \Eng{common notion} (\Gk{koin`h d'oxa}). His writing is elliptical, in the style of lecture-notes---which is probably just what his works are \cite[pp.~xxv \& xxxi]{Aristotle-XVII}; I translate accordingly below, with seemingly missing words supplied in brackets. (Some of the original Greek words in parentheses are the sources of modern technical terms.) In Book \Gk{B} (that is, Book Beta, also called Book III) of the \emph{Metaphysics}, Aristotle introduces some questions: \begin{quote} [996 b 26] Yet indeed, concerning the demonstrative (\Gk{>apodeiktik'oc}) principles, whether they belong to one science (\Gk{>epist'hmh}) or more (\Gk{plei'wn}) is debatable. I call \Eng{demonstrative} the common notions from which everybody proves (\Gk{de'iknumi}) [propositions], for example, \Eng{it is necessary to affirm or deny everything,\footnote{\Gk{p~an >anagka~ion <`h f'anai <`h >apof'anai}.}} or \Eng{it is impossible to be and not be at the same time,\footnote{\Gk{>ad'unaton <'ama e>~inai ka`i m`h e>~inai}.}} and however many other such premisses (\Gk{prot'asic}). \end{quote} Aristotle's examples of common notions are called the Law of the \textsl{Excluded Middle}% \tindexsub{law}{L--- of the Excluded Middle}% \tindexsub{excluded}{Law of the E--- Middle}% \tindexsub{middle}{Law of the Excluded M---} and the Law of \textsl{Contradiction;}% \tindexsub{law}{L--- of Contradiction}% \tindexsub{contradiction}{Law of C---} they are discussed further in Book \Gk{G} (IV). That book opens with a statement of the general subject, which we call \textbf{metaphysics,}% \dindex{metaphysics} but Aristotle called \textbf{first philosophy:}% \dindexsub{first}{--- philosophy}% \dindexsub{philosophy}{first ---, metaphysics}% \label{look-at} \begin{quote} [1003 a 20] There is a science (\Gk{>epist'hmh}) that looks at (\Gk{jewr'ew}) being as such (\Gk{t`o >`on <~h| >'on}) and what applies to it (\Gk{t`a to'utw| 'Estin >epist'hmh tis <'h jewre~i t`o >`on <~h| >'on ka`i t`a to'utw| 'on} (stem \Gk{>'ont-}) is the neuter participle corresponding to the Engish \Eng{being} and the Turkish \Tur{olan}; it appears in modern technical terms like \Eng{ontology}. The feminine stem of the participle is \Gk{o>'us-}; from this is derived the abstract noun \Gk{o>us'ia}, which I translate below as \Eng{beingness}, although a traditional (and misleading) translation is \Eng{substance}.} \end{quote} A Turkish version of this passage, from \cite{Metafizik}, is \begin{quote} Varl\i k olmak bak\i m\i ndan varli\u g\i\ ve ona \"oz\"u gere\u gi ait olan ana nitelikleri inceleyen bir bilim vard\i r. \end{quote} Later in Book \Gk{G}, in ch.~3, Aristotle takes up axioms; but he understands them as something more general than the subject of a particular field like mathematics or physics. First he seems to repeat the question raised in Book~\Gk{B}: \begin{quotation} [1005 a 19] It must be said whether [the inquiry] concerning the so-called axioms (\Gk{>axi'wmata}) of mathematics, and concerning beingness (\Gk{us'ia}), belongs to one science (\Gk{>epist'hmh}) or another (\Gk{apode'ixeic}). So, because it is clear (\Gk{dhl'on}) that [the axioms] apply to all things as beings---for this [namely, being] is common to them---the theory (\Gk{jewr'ia}) concerning them belongs to those who are gaining knowledge (\Gk{gnwr'izontoi}) concerning being as such. Therefore, none of those making particular investigations (\Gk{oepiskopo'untoi}) tries to say something concerning them, wheth\-er [they] are true or not---not the geometer (\Gk{gewm'etrhs}), not the arithmetician (\Gk{>arijmhtik'oc}). But some of the physicists (\Gk{fusiko~i})\footnote{Aristotle's `physicists' are pre-Socratic philosophers such as Thales of Miletus; they are discussed in Book \Gk{A} of the \emph{Metaphysics}.} [were] doing this appropriately (\Gk{e>ik'otwc}). For, they thought they alone were doing research (\Gk{skop'ew}) concerning all nature (\Gk{anwt'erw}) than the physi\-cist---for nature is [just] some one class of being---the inquiry concerning these would belong to the theoretician (\Gk{jewrhtik'oc}) of generality (\Gk{kaj'olou}) and first [or primary] beingness (\Gk{us'ia}). Physics (\Gk{arqa'i}) of the thing (\Gk{pr'agma}): So that the one [who knows best] being as such [can state] the most certain [principles] of all [things]. This is the philosopher. The most certain principle of all is that about which being mistaken is impossible. \end{quote} This principle is the \textbf{Law of Contradiction,}% \dindexsub{contradiction}{Law of C---}% \dindexsub{law}{L--- of Contradiction} which Aristotle now states more precisely than in Book \Gk B: \begin{quote} [1005 b 19] For the same [\textsl{predicate}% \tindex{predicate}] to apply and not apply at the same time to the same [\textsl{subject}% \tindex{subject}] in the same [respect] is impossible.\footnote{\Gk{t`o g`ar a>ut`o <'ama ad'unaton t>~w| a>ut>~w| ka`i kat`a t`o a>ut'o.}} \end{quote} The grammatical\index{grammar} notions of subject and predicate are discussed briefly in \S~\ref{sect:sets} below; there also the Law of Contradiction will be put to use. Meanwhile, a Turkish rendition of Aristotle's formulation of the Law of Contradiction, again from \cite{Metafizik}, is \begin{quote} Ayn\i\ niteli\u gin, ayn\i\ zamanda, ayn\i\ \"ozneye, ayn\i\ bak\i m\i ndan hem ait olmas\i, hem de olmamas\i\ imk\^ans\i zd\i r. \end{quote} After a long discussion of the Law of Contradiction and those who question it, Aristotle gives the \textbf{Law of the Excluded Middle,}% \dindexsub{law}{L--- of the Excluded Middle}% \dindexsub{excluded}{Law of the E--- Middle}% \dindexsub{middle}{Law of the Excluded M---} again with slightly different wording from Book~\Gk B: \begin{quote} [1011 b 23] Neither does [any]thing admit to being between a contradiction, but it is necessary either to affirm or deny one of one, whatsoever.\footnote{\Gk{>All`a m`hn o>ud`e metax`u >antif'asewc >end'eqetai e>~inai o>uj'en, >all'' >an'agkh >`h f'anai >`h >apof'anai <`en kaj'' antif'asis}] for Aristotle is evidently a pair of propositions, one affirming something, the other denying the same thing. The continuation of this passage is in \S~\ref{sect:p-formulas} below. If we follow Aristotle, it seems that, as mathematicians, we need not concern ourselves with the Laws of Contradiction and the Excluded Middle; we can just accept these principles and use them; it is the philosopher's job to identify and enunciate them. But the logician is also a philosopher. In any case, we shall use these principles explicitly in the next section; but we shall also see an apparent violation of one of them. There we shall also begin to state axioms in our mathematical sense. A \textbf{theorem}\dindex{theorem} today is usually considered just as a \emph{noteworthy} proposition with a proof from axioms. The first example is Theorem~\ref{thm:Russell} in the next section. The word \Eng{theorem} itself comes from the Greek \Gk{je'wrhma}, and it is related to the verb with the meaning of \Eng{look at}. (This verb is found at the beginning of Book~\Gk{G} of the \emph{Metaphysics} as quoted above on p.~\pageref{look-at}.) In former times, finer distinctions were considered. A few centuries after Aristotle, Pappus of Alexandria\footnote{Pappus may have been born during the reign of Theodosius I, 379--395 \textsc{bce}, or he may have flourished earlier, during the reign of Diocletian, 284--305 \textsc{bce}. The possibilities are discussed in \cite[pp.~564--567]{MR13:419b}, where also are found the text and translation from which the quotation is adapted.} writes: \begin{quote} Those who favor a more technical terminology in geometrical research use \textbf{problem}\dindex{problem} (\Gk{pr'oblhma}) to mean a [proposition\footnote{Ivor Thomas \cite[p.~567]{MR13:419b} uses \Eng{inquiry} here; but there is \emph{no} word in the Greek original corresponding to this or \Eng{proposition}.}] in which it is proposed to do or construct [something]; and \textbf{theorem,}% \dindex{theorem} a [proposition] in which the consequences and necessary implications of certain hypotheses are investigated; but among the ancients some described them all as problems, some as theorems. \end{quote} A \textbf{lemma}\dindex{lemma} is a proposition proved mainly for the sake of proving other propositions; the first example will be Lemma~\ref{lem:even-odd}. (The Greek \Gk{l'emma} means that which is peeled off, and is from the verb, \Gk{l'epw}, with the meaning of \Eng{peel}.) A \textbf{corollary}\dindex{corollary} to a theorem is a proposition that follows almost immediately from the theorem; the first example will be Corollary~\ref{cor:root-2}. (The word derives from the Latin \Lat{corollarivm,} which is the neuter form of the adjective derived from \Lat{corolla;} this means, among other things, a wreath of flowers \cite{OED}.) \section{Classes, sets, and numbers}\label{sect:sets} In Chapter~\ref{ch:sets}, we shall have a lot to say about \textsl{sets;} but it will be useful to have the basic notion available from the beginning. A \textsl{set}% \tindex{set} is many things, made into one. There are many special cases of sets: Two matching earrings make a \emph{pair;} several football-players make a \emph{team;} the pigeons descending on bread-crumbs in the park make a \emph{flock.} Words like \Eng{pair}, \Eng{team} and \Eng{flock} are \textbf{collective nouns.}% \dindex{collective noun}% \dindexsub{noun}{collective ---}% \dindexsub{word}{noun} In mathematics, the word \Eng{set} is the most general collective noun---except for the word \Eng{class}.\footnote{Levy~\cite{MR1924429} seems to use \Eng{collection} more generally even than \Eng{class}.} In the previous section, I translated Aristotle's word \Gk{g'enoc} as \Eng{class}, but that does not mean that our understanding will be the same as Aristotle's. For us, every set is a {class}, but not every class is a set. Classes and sets are made up of \textbf{elements}\dindex{element} or \textbf{members.}% \dindexsub{member}{--- of a class} In the context of classes, there is no mathematical difference between the words \Eng{element} and \Eng{member}. (However, in an equation, such as~\eqref{eqn:A=B} below, the expressions on either side of the \textsl{sign of equality,}% \tindex{sign of equality}% \tindexsub{equality}{sign of ---} or $=$\glossary{$=$}, can be called the \textbf{members}% \dindexsub{member}{--- of an equation}% \dindexsub{equation}{member of an ---} of the equation.) A \textbf{class}% \dindex{class} is determined by a \textbf{property.}% \dindexsub{proper}{---ty} The property \textbf{defines}% \dindex{defines} the class. I do not attempt to define \Eng{property;} I shall just say that, for every property, there is a class whose members are precisely the things that have that property. This does not mean that a class is necessarily a thing that can itself be a member of classes. Indeed, if we assume that every class \emph{can} be a member of classes, then we can derive a contradiction. This is what we do in the proof of Theorem~\ref{thm:Russell} below. The contradiction is the reason why we have to distinguish classes and sets. A \textbf{set}\dindex{set} is a class that \emph{is} a member of some classes. If $A$ is a set, and $\class C$ is a class, then the sentence \begin{center} \Eng{$A$ is a member of $\class C$} \end{center} is true or false---it is a proposition. Again, all sets are classes; but Theorem~\ref{thm:Russell} shows that not all classes are sets. A class can be indicated in writing or print by the presence of \textbf{braces}\dindex{brace} around its members. So, if we have, say, three objects, \begin{equation*} a,b,c, \end{equation*} then we can form the \emph{single} object \begin{equation*} \{a,b,c\}.\glossary{$\{a,b,c\}$} \end{equation*} This single object is a \emph{class,} namely the class of all things with the property of being one of $a$, $b$, and $c$. In fact, this class will be a \emph{set}. In particular, this set \textbf{contains}\dindex{contains} $a$, $b$, and $c$ (and nothing else) as elements. Elements of a class are \textbf{in}\dindex{in} the class. If $\class C$ is a class, then we have several ways of saying the same thing: \begin{gather*} \Eng{$\class C$ contains $d$;}\\ \Eng{$d$ is an element of $\class C$;}\\ \Eng{$d$ is a member of $\class C$;}\\ \Eng{$d$ is in $\class C$.} \end{gather*} Any of these can be expressed by the symbolism\footnote{The sign $\in$ is apparently derived from the Greek \Gk e. Peano \cite{Peano} used this letter in 1889 as a symbol with the meaning of \Eng{is,} perhaps because the Greek word for \Eng{is} is \Gk{>est'i}. For Peano, $d\in\class C$ means $d$ is a $\class C$, that is, $d$ is one of the things denoted by the term $\class C$.} \begin{equation*} d\in \class C. \end{equation*}\glossary{$d\in\class C$} To \emph{deny} that $d\in \class C$, we can write \begin{equation*} d\notin\class C, \end{equation*}\glossary{$d\notin\class C$} which can be read as \Eng{$d$ is not in $\class C$.} One can say that a class \textbf{comprises}\dindex{comprise}\dindexsub{verb}{comprise} its elements, and the elements \textbf{compose}\dindex{compose}\dindexsub{verb}{compose} the class. Unfortunately, the verbs \Eng{comprise} and \Eng{compose} are often confused by native English-speakers. Alternatively, a set \textbf{consists of}\dindex{consists of} its elements. Words like \Eng{collection}, \Eng{aggregate} and \Eng{family} are sometimes used as synonyms for \Eng{set} (or perhaps for \Eng{class}). I say that a set is \emph{many} things, made into one; but I am using the word \Eng{many} more generally than is usual in ordinary language. A set might have two elements or one element. A set might have \emph{no} element at all: such a set is \begin{equation*} \emptyset, \end{equation*}\glossary{$\emptyset$} the \textbf{empty set}\dindex{empty set}\dindexsub{set}{empty ---}. I shall also assume that sets can have \textsl{infinitely}\tindex{infinite} many elements, and that, in particular, the \textsl{natural numbers}% \tindex{natural number}% \tindexsub{number}{natural ---} compose such a set, namely \begin{equation*} \{0,1,2,3,4,\dots\}. \end{equation*} In Chapter~\ref{ch:numbers}, this assumption will turn out to be a \emph{consequence} of the Axiom of Infinity, \ref{ax:infinity}. A class $\class C$ is \textbf{included in}\dindex{included in} a class $\class D$ if every element of $\class C$ is an element of $\class D$. In this case, we can write \begin{equation*} \class C\included \class D, \end{equation*}\glossary{$\class C\included\class D$} and we can say also\footnote{Some people would say here that $\class D$ \emph{contains} $\class C$; but it is desirable to read $\class C\included \class D$ differently from $c\in \class D$.} that $\class D$ \textbf{includes}\dindex{includes} $\class C$ or that $\class C$ is a \textbf{subclass}\dindexsub{sub}{---class}\dindexsub{class}{sub---} of $\class D$. A subclass of $\class D$ that is also a set is a \textbf{subset}\dindexsub{sub}{---set}\dindexsub{set}{sub---} of $\class D$. If $\class C$ is \emph{not} a subclass of $\class D$, we can write \begin{equation*} \class C\nincluded\class D. \end{equation*}\glossary{$\class C\nincluded\class D$} The first \textsl{axiom} of set-theory is that sets are determined by their elements, so that if two sets have the same elements, then the sets themselves are the same, that is, \textbf{equal.}\dindex{equal} We can express this more symbolically: \begin{axiom}[Extension]\dindexsub{extension}{Axiom of E---}\dindexsub{axiom}{A--- of Extension} \label{axiom:extension} If $A$ and $B$ are sets such that $A\included B$ and $B\included A$, then \begin{equation}\label{eqn:A=B} A=B. \end{equation}\glossary{$A=B$} \end{axiom} Instead of $A\included B$, some people write \begin{equation*} A\pincluded B; \end{equation*}\glossary{$A\pincluded B$} but I prefer to use this to mean that $A$ is a \textbf{proper}\dindexsub{proper}{--- subset}\dindexsub{sub}{proper sub---}\dindexsub{set}{proper sub---} subset of $B$, that is, $A\included B$, but $A\neq B$.\glossary{$A\neq B$} Again, a class is defined by a property. A property can be symbolized by a \textbf{predicate.}\dindex{predicate} A predicate \emph{says something} about a \textsl{subject.}\tindex{subject} (See the Law of Contradiction, in the previous section, as translated from Aristotle.\footnote{The English \Eng{predicate} is from the Latin \Lat{praedicatvm,} a participle of the verb \Lat{praedicare.} This verb consists of the prefix \Lat{prae-} (or \Lat{pre-}) and the verb \Lat{dicare.} This verb, with root \Lat{dic-}, means \emph{say,} and it can be found in various English words, such as \Eng{indicate} and \Eng{dictionary.} The Latin \Lat{praedicatvm} is a translation of the Greek \Gk{kathgore'umenon} \cite{COD6}, a participle of \Gk{kathgor'ew}; this verb consists of the prefix \Gk{kata-} and the verb \Gk{agore'uw}, which means \emph{speak before an assembly of the people;} such an assembly is an \Gk{agor'a}. See Appendix~\ref{Aristotle}. The verb \Gk{kathgor'ew} (or some related word) is the source of the English \emph{category.}}) If $P$ is a predicate, then the corresponding class can be denoted by \begin{equation}\label{eqn:Px} \{x:Px\}; \end{equation}\glossary{$\{x:Px\}$} this is read as \Eng{the class of $x$ such that $P$ [applies to] $x$}; here, the \textsl{variable}% \tindex{variable} $x$ takes the place of a grammatical subject of $P$. If $A$ is a set, then \emph{being an element of $A$} is a property; the class defined by this property is \begin{equation*} \{x\colon x\in A\}. \end{equation*} This class is just the set $A$. Again, by the Extension Axiom, two sets are equal if they have the same members; more generally; two \emph{classes} are equal if they have the same members. In particular, two predicates that are different as symbols may nonetheless define the same class; we may have $\{x\colon Px\}=\{x\colon Qx\}$, even though $P$ and $Q$ are different predicates. Often, in place of $Px$ in~\eqref{eqn:Px}, we have to write something that features $x$ more than once. For example, there is a property of \emph{not being a member of oneself}. In words, the corresponding predicate is something like \begin{equation}\label{eqn:word-predicate} \text{\Eng{\underline{\qquad} is not a member of \underline{\qquad}-self},} \end{equation} with two spaces left for a subject. The phrase \Eng{is not a member of} is also symbolized by $\notin$; so the given property determines the class \begin{equation}\label{eqn:Russell} \{x:x\notin x\}. \end{equation} This is the historically first\footnote{Russell gives the example in a letter \cite{Russell-letter} to Frege in 1902; but Levy \cite[p.~6 correction]{MR1924429} cites an article attributing an independent discovery of the example to Zermelo.} example of a class that is not a set: \begin{theorem}[Russell Paradox]\label{thm:Russell} The class $\{x:x\notin x\}$ is not a set. \end{theorem} \begin{proof}\label{first-proof} Call this class $\Russell$, and suppose it \emph{is} a set. Then by the Law of the Excluded Middle, either $\Russell\in\Russell$ or $\Russell\notin\Russell$. Suppose $\Russell\in\Russell$. Then, by the Law of Contradiction, it is not the case that $\Russell\notin\Russell$. This means $\Russell$ fails to have the defining property of members of $\Russell$, and so $\Russell\notin \Russell$. In short, if $\Russell\in\Russell$, then $\Russell\notin\Russell$. On the other hand, trivially, if $\Russell\notin\Russell$, then $\Russell\notin\Russell$. Having considered both possibilities, we conclude $\Russell\notin\Russell$. This means $\Russell$ \emph{does} have the defining property of members of $\Russell$, so $\Russell\in\Russell$. Thus $\Russell$ is and is not a member of itself. This violates the Law of Contradiction. Therefore the assumption that $\Russell$ is a set must be mistaken, so $\Russell$ is not a set (by the Law of the Excluded Middle). \end{proof} The proof ends with a box,\footnote{Other writers use a different symbol, or none at all. An old-fashioned termination of a proof is \Lat{qed}, for the Latin \Lat{qvod erat demonstrandvm}, with the meaning of \Eng{which was to be demonstrated.}} as noted on p.~\pageref{page:box}. This particular proof is a \textbf{proof by contradiction,}\dindexsub{proof}{--- by contradiction}\dindexsub{contradiction}{proof by ---} because it assumes the falsity of what is to be proved, and it derives from this a violation of the Law of Contradiction. This particular proof also shows that there is a class to which the predicate in~\eqref{eqn:word-predicate} neither applies nor fails to apply. So we have an apparent violation of the Law of the Excluded Middle. I would say rather that we have an example of a class that is not a \emph{real thing,}\index{real} so that it is just meaningless to try to apply predicates to it. \emph{Elements} of classes are the real things. We do assume that subclasses of sets are sets:\footnote{The following Axiom of Separation is also called the \textbf{Axiom of Comprehension;}\tindexsub{axiom}{A--- of Comprehension}\tindexsub{comprehension}{Axiom of C---} but I think this term is better reserved for the original, but false, assumption that every property defines a set.} \begin{axiom}[Separation]\dindexsub{separation}{Axiom of S---}\dindexsub{axiom}{A--- of Separation}\label{ax:separation} Suppose $\universe$\glossary{$\mathcal U$} is some set, and $P$ is a predicate. The class of elements \emph{of $\universe$} to which $P$ applies is a set. \end{axiom} The set created by the axiom is denoted by \begin{equation*} \{x\in\universe\colon Px\}. \end{equation*} I use the letter $\universe$\index{universe}\index{set!universal ---} here because it stands for \Eng{universe;} but the set could be anything. For a mundane example, we could let $\universe$ be the set of human beings living now, and let $P$ be the predicate \Eng{is over two meters tall.} Then $\{x\in\universe\colon Px\}$ is the set of people now taller than two meters. However, in using sets for mathematics, we have no need to consider classes that contain anything other than sets. This is because, by starting with the empty set, and by putting sets into other sets, we can create the \textsl{natural numbers;} from these, we can create all of the other objects of mathematics. The procedure is as follows. Any two classes $\class C$ and $\class D$ have a \textbf{union,}\dindex{union} which is the class comprising every element of $\class C$ or $\class D$ (or both); this union is denoted by \begin{equation*} \class C\cup\class D.\glossary{$\class C\cup\class D$} \end{equation*} (See \S~\ref{sect:Boole}.) \begin{axiom}[Adjunction\footnote{The terminology is from George Boolos, according to Wikipedia \url{http://en.wikipedia.org/wiki/General_set_theory} (accessed September 15, 2010).}]\label{ax:adjunction}% \dindex{Adjunction Axiom} If $A$ is a set, then for all $b$, there is a set whose elements are just $b$ and the elements of $A$. \end{axiom} The new set guaranteed by the Axiom is denoted by \begin{equation*} A\cup\{b\}. \end{equation*} \begin{theorem}[Pairing]\label{thm:pairing} For every $a$ and $b$, the classes $\{a\}$ and $\{a,b\}$ are sets. \end{theorem} \begin{proof} By the Adjunction Axiom, the class $\emptyset\cup\{a\}$ is a set; but this set is just $\{a\}$. Then, by the Axiom again, $\{a\}\cup\{b\}$ is a set; but this set is just $\{a,b\}$. \end{proof} The set $\{a\}$ is called a \textbf{singleton.}\dindex{singleton}\dindexsub{set}{singleton} Evidently the proof can be continued to show that $\{a,b,c\}$ is a set, $\{a,b,c,d\}$ is a set, and so on; in short, \emph{finite}\index{finite} classes are sets. From any set $A$, we can now form the union \begin{equation*} A\cup\{A\}. \end{equation*} This idea can be used to give the following \textbf{recursive definition}% \dindexsub{definition}{recursive ---}% \dindexsub{recursive}{--- definition} of the \textbf{natural numbers.}% \dindex{natural number}% \dindexsub{number}{natural ---} First, we declare that the number \textbf{zero} is just the empty set: \begin{equation*} 0=\emptyset.\glossary{$0$} \end{equation*} Then we define the natural numbers by two rules: \begin{compactenum} \item $0$ is a natural number. \item if $n$ is a natural number, then $n\cup\{n\}$ is a natural number. \end{compactenum} The latter rule assumes that $n$ is a set; but then $n\cup\{n\}$ is also a set. The latter set can be called the \textbf{successor}\dindex{successor} of $n$. Hence all natural numbers are sets, and every natural number has a successor, which is a natural number. To the recursive defition of natural numbers, some writers might add a third condition: \begin{compactenum} \setcounter{enumi}2 \item Nothing else is a natural number. \end{compactenum} However, I understand such a condition to be implicit in every recursive definition as such. If $n$ is a natural number, let us denote its successor $n\cup\{n\}$ by \begin{equation*} \vscr n. \end{equation*} Then we have \begin{align}\label{eqn:A-in-A'} n&\in\vscr n,&n&\included\vscr n. \end{align} The recursive definition of the natural numbers makes \textbf{proof by induction}\dindexsub{proof}{--- by induction}\dindex{induction} in the following sense possible. Suppose $P$ names a property that some natural numbers may have, and suppose moreover that we can establish the following two conditions. \begin{compactenum} \item $P0$. \item For every natural number $n$, if $Pn$, then $P(\vscr n)$. \end{compactenum} Then we have proved by induction that every natural number must have the property (named by) $P$. In the second condition, $Pn$ is called the \textbf{inductive hypothesis.}\dindexsub{inductive}{--- hypothesis}\dindexsub{hypothesis}{inductive ---} The method of proof by induction is first used in Lemma~\ref{lem:includes-elements} below. In general, an inductive proof consists of two steps: \begin{compactenum}[1)] \item the \textbf{base step,}\dindex{base step}\dindexsub{step}{base ---} in which $P0$ is proved; \item the \textbf{inductive step,}\dindexsub{inductive}{--- step}\dindexsub{step}{inductive ---} in which $P(\vscr n)$ is proved from the inductive hypothesis $Pn$. \end{compactenum} It is not obvious that there is even a \textbf{class} consisting of the natural numbers: what \emph{property} do these numbers share? Well, they share the property that they can be obtained by starting with $\emptyset$ and taking successors; but it is not obvious how to make this property precise. One way that works is the following, as we shall show in \S~\ref{sect:ordinals}. We can first define an \textbf{ordinal}% \dindex{ordinal} to be a set $\alpha$ such that \begin{compactenum}[1)] \item $\alpha$ \emph{includes} each of its elements (that is, if $x\in \alpha$, then $x\subseteq \alpha$); \item if $\alpha$ has two distinct elements, then one of them contains the other; \item If $A\included \alpha$, and $A$ is not empty, then $A$ has an element $b$ that is contained by all of the other elements of $A$, though not by $b$ itself. \end{compactenum} Then every element of an ordinal is an ordinal. An ordinal is a \textbf{limit}\dindex{limit} if it is not empty and is not of the form $\beta\cup\{\beta\}$ for any set $\beta$. Then a natural number is an ordinal that neither \emph{is} a limit nor \emph{contains} limits. Again, this will be worked out in \S~\ref{sect:ordinals}; meanwhile, let us accept the informal definition of the natural numbers. The class of natural numbers is denoted by \begin{equation*} \vnn. \end{equation*} This symbol is not the Latin minuscule letter \letter w (the so-called double \letter u); it is the Greek minuscule \emph{omega.} Observe that \Eng{mega} means big, so an omega is a big \letter o---rather, a double \letter o, or \Eng{oo}, which, if written quickly, may come out looking like $\vnn$. As we have just defined them, the natural numbers can be called more precisely the \textbf{von-Neumann natural numbers.}% \dindexsub{number}{non-Neumann natural ---}% \dindexsub{natural number}{von-Neumann ---}% \dindexsub{von Neumann}{--- natural number}% \footnote{These natural numbers are instances of the von-Neumann \textsl{ordinal numbers,}% %\tindex{ordinal}% \tindexsub{number}{ordinal ---} defined by von Neumann in 1923 \cite{von-Neumann}. However, in his introduction to von Neumann's paper, van Heijenoort cites Bernays as saying that Zermelo had a similar idea for ordinals in 1915; also, in this context, Levy \cite[II.3.8, p.~52]{MR1924429} cites Zermelo from 1916.} The first four von-Neumann natural numbers are \begin{align*} &\emptyset,& &\{\emptyset\},& &\bigl\{\emptyset,\{\emptyset\}\bigr\},& &\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}, %\{\emptyset, \{\emptyset\}, \{\emptyset,\{\emptyset\}\}, % \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}\}, \end{align*} where again $\emptyset=0$. We have the following standard symbols for some successors: \begin{equation*} \begin{array}{c||c|c|c|c|c|c|c|c|c} n & 0&1&2&3&4&5&6&7&8\\ \hline \vscr n &1&2&3&4&5&6&7&8&9 \end{array} \end{equation*} Also, we may write \begin{equation*} n+1 \end{equation*}\glossary{$n+1$} for $\vscr n$. If $m$ and $n$ are in $\vnn$, and $m\included n$, then we usually write \begin{equation*} m\leq n. \end{equation*}\glossary{$m\leq n$} The class $\vnn$ has two more properties, besides admitting proofs by induction; these are given by the next two theorems. \begin{theorem}\label{thm:z} $0$ is not the successor of any natural number. \end{theorem} \begin{proof} We argue by contradiction. Suppose $0$ is a successor; say $0=\vscr n$. But $n\in\vscr n$, as noted in~\eqref{eqn:A-in-A'}; so $n\in 0$. This contradicts that $0$ is empty. Therefore $0$ is not a successor. \end{proof} \begin{lemma}\label{lem:includes-elements} Every von-Neumann natural number includes all of its elements. \end{lemma} \begin{proof} We use induction. Let $P$ be the predicate \begin{center} \Eng{\underline{\qquad} includes all of its elements.} \end{center} Since $0$ has no elements, trivially $0$ includes all of its elements. Therefore $P0$. This completes the base step of the proof. For the inductive step, suppose $Pn$ (as an inductive hypothesis). Say $k\in \vscr n$. Since $\vscr n=n\cup\{n\}$, either $k\in n$, or $k\in\{n\}$. If $k\in n$, then $k\included n$ by inductive hypothesis. If $k\in\{n\}$, then $k=n$, so $k\included n$. In either case, $k\included n$. But $n\included \vscr n$. Hence $k\included\vscr n$. (We use the obvious proposition that if $A\included B$ and $B\included C$, then $A\included C$; this proposition will be part of Lemma~\ref{lem:implications}.) In short, if $k\in\vscr n$, then $k\included\vscr n$. Therefore $P(\vscr n)$. This completes the induction. \end{proof} \begin{theorem}\label{thm:u} Natural numbers with the same successor are the same. \end{theorem} \begin{proof} Suppose $k$ and $n$ are natural numbers, and $\vscr k=\vscr n$. We must show $k=n$. We have \begin{equation*} k\cup\{k\}=n\cup\{n\}. \end{equation*} In particular, $k\in n\cup\{n\}$ and $n\in k\cup\{k\}$. Suppose if possible $k\neq n$. Then we must have $k\in n$ and $n\in k$, hence $k\included n$ and $n\included k$ by the previous lemma, and therefore $k=n$ by the Axiom of Extension,~\ref{axiom:extension}. This contradicts the assumption that $k\neq n$; therefore the assumption is false, and $k=n$. \end{proof} We can call $m$ an \textbf{immediate predecessor}\dindex{immediate predecessor}\dindexsub{predecessor}{immediate ---} of $\vscr m$. By our recursive definition, every natural number that is not $0$ must be $\vscr m$ for some $m$; that is, every natural number $n$ other than $0$ has an immediate predecessor. By the last theorem, this predecessor is \emph{unique:}\index{unique} it is \emph{the} immediate predecessor of $n$, and it can be denoted by \begin{equation*} n-1.\glossary{$n-1$} \end{equation*} The von-Neumann definition of the natural numbers is convenient, because according to this definition, each natural number $n$ is just the set that can be denoted by \begin{equation*} \{0,\dots,n-1\}. \end{equation*}\glossary{$\{0,\dots,n-1\}$} If $n=0$, then this is the empty set. If we do not happen to care about whether each natural number is a particular set, then we can denote the set of natural numbers by \begin{equation*} \N;\glossary{$\mathbb N$} \end{equation*} this is the usual notation when one is not interested in set-theory. Then $\N$ is just a class that contains an element $0$, and whose every element $n$ has a successor, which can be denoted by \begin{equation}\label{eqn:scrn} \scr n\glossary{$\scr n$} \end{equation} or $n+1$\glossary{$n+1$}, such that: \begin{compactenum}[1)] \item $0$ is not the successor of any element of $\N$; \item elements of $\N$ with the same successor are the same; \item $\N$ is included in every class that contains $0$ and that, for every $n$ in $\N$, contains $\scr n$ if it contains $n$. \end{compactenum} These conditions on $\N$ are the \textsl{Peano Axioms;}\tindexsub{Peano}{--- Axioms}\tindexsub{axiom}{Peano A---s} we shall show in Chapter~\ref{ch:numbers} that all properties of $\N$ follow from them. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove by induction that every element of $\vnn$ either \emph{is} $0$ or \emph{contains} $0$. \item What is wrong with the following proof that every element of $\vnn$ is equal to each of its elements? \begin{quote} For all $n$ in $\vnn$, if $k\in n$, we show $k=n$. We use induction on $n$. The claim is trivially true when $n=0$, since $0$ has no elements. Suppose the claim is true when $n=m$. Suppose $k\in\vscr m$. Then either $k\in m$ or $k=m$. If $k\in m$, then by inductive hypothesis, $k=m$. Therefore, in any case, $k=m$. That is, every element of $\vscr m$ is $m$. But by inductive hypothesis, $m$ is equal to its immediate predecessor (since this is an element of $m$). Let the immediate predecessor of $m$ be $\ell$. Then $\ell=m$, so $\vscr{\ell}=\vscr m$. But $\vscr{\ell}=m$, since $\ell$ is the immediate predecessor of $m$. Therefore $m=\vscr m$. If $k\in\vscr m$, we have already shown $k=m$; now we can conclude $k=\vscr m$. This completes the induction. \end{quote} \item From the `theorem' in the preceeding exercise, prove that all natural numbers are equal to $0$. \end{enumerate} \section{Algebra of the integers}\label{algebra} Now that we have, from the previous section, a precise definition of the natural numbers, I want to review some things that we know about them from school. We cannot yet define all of these things precisely, or prove them: this will happen in Chapter~\ref{ch:numbers}. Meanwhile, we just have a set called $\N$, whose members form the list \begin{equation*} 0,1,2,3,\dots. \end{equation*} As we have seen, every natural number $n$ has a successor, which is usually denoted by $n+1$. Some mathematicians start the list of natural numbers at $1$ instead of $0$; but I shall just say that the members of the set $\{1,2,3,\dots\}$ are the \textbf{positive}\dindexsub{number}{positive ---}\dindexsub{positive}{--- number} natural numbers. The number $0$ does not have an immediate predecessor that is a natural number; but it does have the immediate predecessor called $-1$.\glossary{$-1$} This is not a natural number, but it is an \textsl{integer}. The set of \textbf{integers}\dindex{integer}\dindexsub{number}{integer} comprises every natural number, along with a \textbf{negative,}% \dindex{negative} denoted by $-n$,\glossary{$-n$} for each positive natural number $n$. Then $-n$ has the successor $-(n-1)$ and the immediate predecessor $-(n+1)$. The integers that are not natural numbers are also called \textbf{negative}% \dindexsub{integer}{negative ---}% \dindexsub{number}{negative integer}% \dindexsub{negative}{--- integer} integers. \emph{Every} integer $n$ has a \textbf{negative,} denoted by $-n$, although this number is itself negative only if $n$ is positive. The set of integers is commonly denoted by\footnote{Here the letter zed or zee stands for the German $\mathfrak{Zahl}$, \Eng{number.} In English, the integers are also called \textbf{whole numbers.}\dindex{whole number}\tindexsub{number}{whole ---} In fact, the English word \Eng{integer} comes from the Latin \Lat{integer,} which means whole. This Latin word developed in France into the French word \textsf{entier,} which entered English and became \Eng{entire}. Thus two English words---\Eng{integer} and \Eng{entire}---represent the same Latin word. People interested in such matters may refer to such pairs of words as \textsl{doublets.}\tindexsub{double}{---t}\tindexsub{word}{doublet}} \begin{equation*} \Z.\glossary{$\mathbb Z$} \end{equation*} This set is equipped with three \textsl{operations,}\tindex{operation} namely \textbf{addition,}\dindex{addition}\dindexsub{operation}{addition} \textbf{additive inversion,}\dindexsub{additive}{--- inversion}\dindexsub{inversion}{additive ---}\dindexsub{operation}{additive inversion} and \textbf{multiplication.}\dindex{multiplication}\dindexsub{operation}{multiplication} (Operations are \textsl{functions;}\tindex{function} functions in general and operations in particular are defined formally in \S~\ref{sect:functions}.) In particular, if $x$ and~$y$ are integers, then so are \begin{compactenum}[1)] \item $x+y$\glossary{$x+y$} (the \textbf{sum}\dindex{sum} of $x$ and $y$, which here are \textbf{addends}\dindex{addend}), \item $-x$\glossary{$-x$} (\textbf{minus-}$x$\dindex{minus}, the \textbf{additive inverse}\dindexsub{additive}{--- inverse}\dindexsub{inverse}{additive ---} or \textbf{negative}% \dindex{negative} of $x$), and \item $x\cdot y$\glossary{$x\cdot y$} (the \textbf{product}\dindex{product} of the \textbf{factors}\dindex{factor} $x$ and $y$). \end{compactenum} By convention, multiplication is also indicated by \textbf{juxtaposition;}\dindex{juxtaposition} that is, the product $x\cdot y$ is also denoted by \begin{equation*} xy.\glossary{$xy$} \end{equation*} Something like the symbol for additive inversion is also used for a fourth operation, \textbf{subtraction,}\dindex{subtraction} which can be defined in terms of the other operations. \emph{Subtracting}\annot{The English verb \Eng{subtract} is sometimes pronounced as if it were \Eng{substract}. The English verb comes from a participle of the Latin verb whose infinitive is \Lat{subtrahere}. This verb is in turn built up from \Lat{trahere} (meaning draw or carry) and the preposition \Lat{sub} (meaning from below or away). According to the OED \cite{OED}, in medieval times, an \letter s was inserted between \Lat{sub} and \Lat{trahere}, yielding \Lat{substrahere}, from which came \Eng{substract} in English; but this formation is considered incorrect. The English word \Eng{abstract} is from the Latin \Lat{abstrahere}, but here the \letter s belongs properly to the preposition \Lat{abs}, although the preposition is more commonly seen as \Lat{ab} or even \Lat{a}.} $y$ from $x$ produces a \textbf{difference,}\dindex{difference} which is denoted by \begin{equation*} x-y, \end{equation*} and which is just the sum of $x$ and $-y$. Note that $x-y$ is not generally the same as $y-x$. If we want to assign names, then, in the difference $x-y$, we can call $x$ the \textbf{minuend}\dindex{minuend} (from the Latin, with the meaning of \Eng{that which is to be diminished}), and we can call $y$ the \textbf{subtrahend}\dindex{subtrahend} (\Eng{that which is to be subtracted}). Subtraction is thus a \textbf{composition}\dindex{composition} of two other operations. The process of computing $x-y$ can be indicated by a \textbf{tree,}\dindex{tree}\annot{Trees as such are covered in a later course, Math 112.} thus: %\input{first-trees.tex} \begin{equation*} \xymatrix@!0{ *+[F]{x} & & & *+[F]{y} \\ & & *+[o][F]{-} \ar@{-}[ur] &\\ & *+[o][F]{+} \ar@{-}[uul] \ar@{-}[ur] & & } \end{equation*} More complicated compositions and trees are possible. For example, the tree \begin{equation*} \xymatrix@!0{ *+[F]{x} & & & *+[F]{y} & & *+[F]{z} & & *+[F]{w}\\ && *+[o][F]{-} \ar@{-}[ur] &&&& *+[o][F]{+} \ar@{-}[ul] \ar@{-}[ur] & \\ &&&& *+[o][F]{\cdot} \ar@{-}[ull] \ar@{-}[urr] &&&\\ & *+[o][F]{+} \ar@{-}[uuul] \ar@{-}[urrr] &&&&&& } \end{equation*} indicates the sum of $x$ and the product of minus-$y$ and the sum of $z$ and $w$. Usually this sum is written on one line, as \begin{equation}\label{eqn:arith-term} x+-y\cdot(z+w), \end{equation} or more simply as \begin{equation*} x-y(z+w). \end{equation*} I shall refer to such a \textbf{string}\dindex{string} of symbols as an \textsl{arithmetic term.}% %\tindexsub{term}{arithmetic ---}\tindexsub{arithmetic}{--- term} \footnote{Here the word \Eng{arithmetic} is an adjective and is pronounced with the stress on the penultimate (next-to-last) syllable.} (The Greek word\footnote{Strictly, the Greek word \Gk{>arijm'oc} refers to \emph{a number of things}, in particular, more than one;---certainly not zero or `fewer' than zero. See \cite{MR1215482}.} for \Eng{number} is \Gk{>arijm'oc}, which is \Lat{arithmos} in Latin letters. Our general definition\footnote{In another context, Aristotle's definition of \Eng{term} is in Appendix~\ref{Aristotle}.} of \Eng{term} comes in \S~\ref{sect:1st}.) Officially, \textbf{(arithmetic) terms}\dindexsub{arithmetic}{--- term}\dindexsub{term}{arithmetic ---} will be certain strings composed of \begin{compactenum}[1)] \item the symbols $+$, $-$ and ${}\mathrel{\cdot}{}$ (a dot); \item \textbf{variables,}\dindex{variable} such\footnote{The convention of using letters from the end of the Latin alphabet for `unknown quantities' dates back to Descartes; see \cite{Descartes-Geometry}. Since we don't want any limit on the number of variables we can use, and yet we want to define things precisely, we could declare officially that our variables must come from the list $x$, $x'$, $x''$, and so forth.} as $x$, $y$ and $z$; \item symbols for certain integers, such as $12$, $0$ and $-137$---such symbols can be called \textbf{numerals}\dindex{numeral}\footnote{It is probably simplest to think of a numeral as a single symbol, even though, typographically, it may be a string of digits, possibly preceeded by a minus-sign. For example, the numeral $-137$ should be thought of as a single symbol like $c_{-137}$ (that's $c$ with the subscript~$-137$). Our decimal convention for writing numerals is just that, a convention; it has no essential relation to our definition of arithmetic terms. See also Footnote~\ref{footnote:+} below.} or \textbf{(numeral) constants}% \dindex{constant}% \dindexsub{constant}{numeral ---}% \footnote{Letters from the front of the Latin alphabet are used to denote such constants; again the convention is found in Descartes. Used in this way, the letters can be called \textbf{literal constants,}\dindex{literal constant}\dindexsub{constant}{literal ---} where the word \Eng{literal} is just the adjectival form of \Eng{letter}. But for us, literal constants are not \emph{literally} parts of terms; they just \emph{stand} for parts of terms---namely, numerals.}; \item the parentheses $($ and $)$.\glossary{$(\quad)$} \end{compactenum} The formal definition of arithmetic terms is {recursive}\index{recursive!--- definition}\index{definition!recursive ---}, in the sense of the previous section: \begin{compactenum} \item Every variable is an arithmetic term. \item Every numeral is an arithmetic term. \item If $t$ is an arithmetic term, then so is $-t$. \item If $t_0$ and $t_1$ are arithmetic terms, then so are $(t_0+t_1)$ and $(t_0\cdot t_1)$. \end{compactenum} Our definition of arithmetic terms is recursive in the following way. Suppose $A$ is \emph{some} set of strings of symbols such that each of the following conditions holds: \begin{compactenum} \item Every variable is in $A$. \item Every numeral is in $A$. \item If $t$ is in $A$, then $-t$ is in $A$. \item If $t_0$ and $t_1$ are in $A$, then so are $(t_0+t_1)$ and $(t_0\cdot t_1)$. \end{compactenum} Then $A$ contains all arithmetic terms. Therefore, {proof by induction}\index{proof!--- by induction}\index{induction!proof by ---} on arithmetic terms is possible; here is an example: \begin{proposition}\label{prop:arith-terms} Every arithmetic term has as many left parentheses as right parentheses. \end{proposition} \begin{proof} Let $A$ be the set of arithmetic terms that have as many left parentheses as right parentheses. Then $A$ contains all variables and constants (since these have no parentheses). Suppose $A$ contains $t$. Then $t$ has as many left as right parentheses (just because it is in $A$), so the same is true of $-t$. This means $-t$ is in $A$. Similarly, if $t_0$ and $t_1$ are in $A$, then each of them has as many left as right parentheses, so the same is true of $(t_0+t_1)$ and $(t_0\cdot t_1)$; \begin{comment} (for example, if $t_0$ has $n_0$ left parentheses, and $t_1$ has $n_1$ left parentheses, then $(t_0+t_1)$ has $n_0+n_1+1$ left parentheses) \end{comment} this means these terms are also in $A$. By the recursive definition of arithmetic terms, every term is in $A$. \end{proof} By the formal definition of arithmetic terms, string~\eqref{eqn:arith-term} above is not strictly a term; to satisfy the definition, the term should be written as \begin{equation*} (x+(-y\cdot(z+w))). \end{equation*} By convention, we can leave out the dot between $-y$ and $(z+w)$, and we can remove some of the parentheses. But we can do this only because we have a conventional \textbf{order of operations}\dindexsub{order}{--- of operations}\dindexsub{operation}{order of ---s} in terms. By this convention, expressions in brackets are evaluated before all else, and then multiplication is performed before addition (and subtraction), but otherwise operations are performed as they are read from left to right. So, $(x+y)z$ means something different from $x+yz$: the former is an informal version of the term $((x+y)\cdot z)$; the latter, of $(x+(y\cdot z))$. The formal definition of arithmetic terms should ensure that each term indicates uniquely how to calculate an integer, once integral values are assigned to the variables. In short, arithmetic terms should be \textbf{uniquely readable.}\dindexsub{unique}{---ly readable}\dindexsub{readable}{uniquely ---} As we have defined them, they \emph{are} uniquely readable: this is a theorem with a proof like that of Theorem~\ref{thm:UR} below. An arithmetic term is not exactly the same thing as a \textsl{polynomial.} For example, the terms $(x\cdot(y+z))$ and $((x\cdot y)+(x\cdot z))$ are different. However, they always yield the same number if $x$, $y$ and $z$ are respectively replaced by the same three integers. We therefore write \begin{equation}\label{eqn:identity} x(y+z)=xy+xz, \end{equation} and we shall say that the two members of this equation \textbf{represent}\dindex{represent} the same \textbf{polynomial.}\dindex{polynomial} Also, Equation (\ref{eqn:identity}) is called an \textbf{(arithmetic) identity.}\dindexsub{arithmetic}{--- identity}\dindexsub{identity}{arithmetic ---} An equation of arithmetic terms can be called a \textbf{Diophantine equation,}% \dindex{Diophantine equation}% \dindexsub{equation}{Diophantine ---} in memory of the ancient Alexandrian mathematician Diophantus, who studied such equations.\footnote{Diophantus wrote the \emph{Arithmetica,} in thirteen books, of which six have come down to us \cite[pp.~516, n.~$a$]{MR13:419b}. One problem that he considers, for example, is, in our notation, to find rational solutions to the pair \begin{gather*} 8x+4=y^2,\\ 6x+4=y^2 \end{gather*} of equations \cite[pp.~526--535]{MR13:419b}.} A Diophantine equation is an example of an \textsl{(arithmetic) formula}\tindexsub{arithmetic}{--- formula}\tindexsub{formula}{arithmetic ---}. For example, the equation \begin{equation}\label{eqn:elliptic} y^2=4x^3-ax-b \end{equation} (where $a$ and $b$ are understood to be integers) is an arithmetic formula. Its \textbf{solutions}\dindex{solution} are those pairs of integers that \textbf{satisfy}\dindex{satisfy} the equation: those pairs $(c,d)$ of integers such that $d^2=4c^3-ac-b$. Formula~\eqref{eqn:elliptic} is not an identity, because not every pair of integers satisfies it. (For example, if $(c,d_0)$ and $(c,d_1)$ satisfy it, then we must have $d_1=\pm d_0$; there is no other possibility.)\footnote{Equations like \eqref{eqn:elliptic} are of ongoing interest to number-theorists. It is a twentieth-century result that the equation $y^2=x^3+17$ has two solutions, $(-2,3)$ and $(2,5)$, from which all rational solutions can be found by certain rules; and only eight of these solutions are integral \cite[Example~III.2.4, pp.~59~f.]{MR817210}.} By our definition, a polynomial is an abstraction from the notion of a term. It is an \textsl{equivalence-class}\tindexsub{equivalence}{---{}-class} of terms, in the sense of \S~\ref{sect:eq}. You can think of a polynomial as an operation. Then a term is a set of instructions---a recipe for how to perform the operation. The point then is that the same operation can be performed in different ways. This is why different terms can represent the same polynomial; this is why we have nontrivial identities like~\eqref{eqn:identity}. For example, the term $x+y$ says, `Start with $x$, and add $y$'; the term $y+x$ says, `To $y$, add $x$.' These are different activities, but they yield the same result; so we write $x+y=y+x$. How can you tell when two terms represent the same polynomial? It is easy to show when they represent different polynomials. For example, $x^2$\glossary{$x^2$} (that is, $xx$) represents a different polynomial from $x$, since $(-1)^2\neq-1$. But how do we know that the two members of Equation~(\ref{eqn:identity}) represent the same polynomial? As an identity, the equation expresses the \textbf{distributive}\dindex{distributive} property of multiplication over addition. So how do we know that multiplication \emph{has} this property with respect to addition? We can check it for certain integers, say $x=5$ and $y=17$ and $z=-14$: \begin{gather*} 5(17+-14)=5\cdot 3=15;\\ 5\cdot 17+5\cdot-14=85-70=15. \end{gather*} But we cannot check the property for all integers in this way, since there are infinitely many integers. Strictly speaking, if one wants to use the distributive property with full understanding, then one should give precise definitions of the integers and their operations, and then one should \textsl{prove}% \tindex{prove} the distributive property. We shall be able to do this in Chapter~\ref{ch:numbers}: see Theorem~\ref{thm:mult}. However, we did not need to know all of the properties like the distributive property, just to be able to \emph{define} the notion of a polynomial. As we have discussed them so far, the integers form the \textsl{structure}% \tindex{structure} \begin{equation}\label{eqn:Z} (\Z,-,+,\cdot). \end{equation} Structures are defined generally in \S\S~\ref{sect:relations} and~\ref{sect:1st}. The structure in~\eqref{eqn:Z} is the set $\Z$ equipped with certain specified operations, namely addition, additive inversion and multiplication. Now, $\Z$ also has the named\footnote{\label{footnote:+}In fact, \emph{every} integer can be given a name in decimal notation. Alternatively we can just write every positive integer as the appropriate sum $1+1+\dots+1$, write zero as $0$, and write every negative integer as $-(1+\dots+1)$.} elements $0$ and $1$. Moreover, $\Z$ is equipped with the \textsl{ordering}\tindexsub{order}{ing} denoted by $<$\glossary{$xy$ is read as \Eng{$x$ is greater than $y$}, and means $y$, $\leq$, and $\geq$. In this context, we may also speak of the \textbf{inequation}\dindex{inequation} \begin{equation*} t_0\neq t_1, \end{equation*} which is satisfied in $\Z$ by just those integers that do \emph{not} satisfy the equation $t_0=t_1$. The positive integers\index{positive!--- integer}\index{integer!positive ---} are just the positive natural numbers; symbolically, these are the integers that satisfy the inequality $0arijmo`i ple'iouc e>is`i pant`oc to~u protej'entoc pl'hjouc pr'wtwn >arijm~wn}: `The prime numbers are more than any given multitude of prime numbers.' If for \Eng{multitude} we understand \Eng{set}, then, for Euclid, there is no such thing as an \emph{infinite} set; in particular, there is no set such as we have called $\N$.} \end{proposition} \begin{proof} Suppose there are only finitely many prime numbers. Then there are $n$ primes for some $n$ in $\N$. We can now list the primes thus: \begin{equation*} p_0,p_1,\dots,p_{n-1}. \end{equation*} The product $p_0p_1\cdots p_{n-1}$ must be divisible by each prime $p_i$ on our list, and therefore the sum \begin{equation*}1+p_0p_1\cdots p_{n-1}\end{equation*} is indivisible by each prime $p_i$ (why?). Therefore this sum has a prime factor not on our list of primes. This contradicts our assumption that our list contains all primes. Therefore there are infinitely many primes. \end{proof} Are you satisfied with the proof of Proposition \ref{thm:inf-primes}? What details does it leave out? We have not proved that every positive integer (besides $1$) \emph{has} prime factors. (However, this fact is Euclid's Proposition VII.32; see also Example~\ref{example:si} below.) Nor have we defined what `infinitely many' means. (We shall in \S~\ref{Peano}.) Still, by some standards, we \emph{have} given a proof: a proof by contradiction.\annot{A proof with a similar level of detail is offered to the general reader by Hardy \cite[\S~12]{MR92j:01070}.} \subsection*{Incommensurability of diagonal and side} The next proposition is also proved by contradiction. We first need a definition and some {lemmas}. An integer is \textbf{even}\dindex{even number}\dindexsub{number}{even ---} if $2$ divides it; otherwise, the integer is \textbf{odd}\dindex{odd number}\dindexsub{number}{odd ---}; so $2n$ is even, but $2n+1$ is odd. \begin{lemma}\label{lem:even-odd} The product of two integers is \begin{compactenum}[1)] \item \emph{even}, if one of the integers is even; \item \emph{odd}, otherwise. \end{compactenum} \end{lemma} \begin{proof} Let the two integers be $a$ and $b$. If $a$ is even, so that $2\divides a$, then $a=2c$ for some integer $c$, so $ab=2cb$, which means $ab$ is even. If $a$ and $b$ are odd, then they are $2c+1$ and $2d+1$ for some integers $c$ and $d$, so that $ab=(2c+1)(2d+1)=4cd+2c+2d+1=2(2cd+c+d)+1$, which is odd. \end{proof} %The proof used standard facts about arithmetic, which however we shall %not be able to state and prove \emph{precisely} until %Chapter~\ref{ch:numbers}. The following is a fundamental property\footnote{Born around 1601, Pierre Fermat developed the method of \textsl{infinite descent}% \tindexsub{infinite}{--- descent}% \tindexsub{descent}{infinite ---}% \dindexsub{method}{--- of infinite descent} to prove such theorems as that no right triangle whose sides are integral has an area that is the \emph{square} of an integer: If there were such a triangle, then there would be a smaller one, and so on. See Weil \cite[II.IX, pp.~75~ff.]{MR734177}.} of $\N$; we shall use it here and there before proving it in Chapter~\ref{ch:numbers}. (It is a consequence of the \textsl{Peano Axioms}\tindexsub{Peano}{--- Axioms}\tindexsub{axiom}{Peano A---s} given at the end of \S~\ref{sect:sets}, but it cannot be proved by induction alone.) \begin{lemma}[Infinite Descent]% \label{lem:inf-desc}% \index{infinite!proof by --- descent}% \index{proof!--- by infinite descent}% \index{descent!proof by infinite ---} Every \textsl{strictly decreasing}% \tindexsub{strict}{---ly decreasing}% \tindexsub{decreasing}{strictly ---} \textsl{sequence}% \tindex{sequence} of positive integers must be finite: that is, if there is a sequence $(a_0,a_1,a_2,a_3, \dots)$ of positive integers such that \begin{equation*} a_0>a_1>a_2>a_3>\dotsb, \end{equation*} then the sequence must stop---must have a final entry $a_n$ for some $n$ in $\N$. \end{lemma} \begin{proof} The claim follows because $\N$ is \textsl{well-ordered}\tindexsub{well}{---{}-ordered}\tindexsub{order}{well-{}---ed}, which means that every non-empty subset of $\N$ has a least element; we shall discuss this in \S~\ref{sect:well-ordered}. The set of terms in a strictly decreasing sequence $(a_0,a_1,\dots)$ of positive integers must have a least element, $a_n$; then there can be no term after this, since it would be less than $a_n$. \end{proof} We can now state and prove the following. Its geometric interpretation is that there is no unit length into which the diagonal and side of a square can be divided. Aristotle\footnote{In the \emph{Prior Analytics;} the passage is quoted and discussed at \cite[pp.~110 f.]{MR13:419a}.} alludes to a proof similar to ours. \begin{proposition}\label{thm:2} The Diophantine equation \begin{equation}\label{eqn:Dioph-2} x^2=2y^2 \end{equation} has no non-zero integral solution. \end{proposition} \begin{proof} Suppose, if possible, that $(a_0,a_1)$ satisfies the equation, where $a_0$ and $a_1$ are non-zero integers. In particular then, \begin{equation}\label{eqn:root-2} {a_0}^2=2{a_1}^2. \end{equation} Hence ${a_0}^2$ is even, so $a_0$ is even by Lemma~\ref{lem:even-odd} (since if $a_0$ were odd, then ${a_0}^2$ would be odd\footnote{Here, in the notation of \S~\ref{sect:entailment}, we use $\sv P\lto\sv Q,\lnot\sv P\lto\lnot\sv Q\models\sv Q\lto\sv P$.}); say $a_0=2a_2$. Then ${a_0}^2=4{a_2}^2$; this, with~\eqref{eqn:root-2}, implies\footnote{The properties of equality that allow this conclusion are discussed in detail in \cite[Ch.~III, pp.~54--67]{Tarski-Intro}.} that $2{a_1}^2=4{a_2}^2$, hence \begin{equation*}\label{eqn:root-2next} {a_1}^2=2{a_2}^2. \end{equation*} Thus $(a_1,a_2)$ is also a solution of~\eqref{eqn:Dioph-2}. In short, given the solution $(a_0,a_1)$, we can find a solution $(a_1,a_2)$. Continuing, we can find an integer $a_3$ such that ${a_2}^2=2{a_3}^2$, and so forth. That is, there is an infinite sequence \begin{equation*} a_0, a_1, a_2, a_3, \dots \end{equation*} of integers $a_k$ such that $(a_k,a_{k+1})$ is a solution of~\eqref{eqn:Dioph-2} for each natural number $k$. (Strictly, the existence of such a sequence is only justified by the Recursion Theorem,\index{recursion!R--- Theorem} which is~\ref{thm:recursion} below.) But we may also assume (why?) that each integer $a_k$ is \emph{positive.} Then \begin{equation*} a_0>a_1>a_2>a_3>\dotsb, \end{equation*} which is absurd: no such sequence can be infinite, by Lemma~\ref{lem:inf-desc}. Therefore such $a_0$ and $a_1$ cannot exist. \end{proof} \subsection*{Euclidean algorithm} An alternative proof of the last proposition is given in \S~\ref{sect:anthyphaeresis} in terms of the \textsl{Euclidean algorithm}\tindex{Euclidean algorithm}\tindexsub{algorithm}{Euclidean ---} for finding the greatest common divisor of two positive integers. Suppose $a$ and $b$ are positive integers. Then there is a unique natural number $k$ such that \begin{equation}\label{eqn:goes-into} ka\leq b<(k+1)a. \end{equation} We say that $k$ is the \textbf{number of times}\dindexsub{number}{--- of times}\dindexsub{times}{number of ---} that $a$ goes into $b$. Then $b-ka$ is the \textbf{remainder}\dindex{remainder} after division of $b$ by $a$. Let us denote this remainder by \begin{equation*} \rem ba. \end{equation*} So we have $b=ka+\rem ba$ for some integer $k$, and $0\leq\rem baarijm'oc}, of the earlier English \Eng{algorism}, which was adapted from \Eng{al-Kow\=arasm\=\i}, the surname of Abu Ja'far Mohammed Ben Musa, whose work in algebra gave the so-called Arabic numerals to Europe.} A modern formulation of this algorithm is found in \cite{Dries-Mosch}: \begin{equation*} \gcd(a,b)= \begin{cases} b,&\text{ if }\rem ab=0;\\ \gcd(b,\rem ab),&\text{ otherwise;} \end{cases} \end{equation*} assuming $00$? \item Supply the missing details of the proof of Proposition~\ref{thm:gcd}; specifically, for all $n$ in $\N$, prove: \begin{enumerate} \renewcommand{\labelenumii}{(\theenumii)} \item $a_{n+1}anjuya'iresic}; see \cite[pp.~504--509]{MR13:419a}.} generating a sequence $a_0$, $a_1$, $a_2$, \dots, possibly finite, of non-negative real numbers, and a corresponding sequence, $n_0$, $n_1$, \dots, of natural numbers. Call the latter sequence the \textbf{anthyphaeretic sequence}\dindexsub{sequence}{anthyphaeretic ---} of $(a_0,a_1)$. Then we have shown that the anthyphaeretic sequence of $(1+\radix 2,1)$ is $2$, $2$, $2$, \dots, never ending. That the Ancients found interest in such sequences can be inferred from certain old texts: see the brief discussion at \cite[pp.~508~f.]{MR13:419a}. In modern notation, we have \begin{gather*} a_k=n_k\cdot a_{k+1}+a_{k+2},\\ \frac{a_k}{a_{k+1}}=n_k+\frac{a_{k+2}}{a_{k+1}}= n_k+\cfrac 1{\Bigl(\cfrac{a_{k+1}}{a_{k+2}}\Bigr)},\\ \frac{a_0}{a_1}=n_0+\cfrac 1{n_1+\cfrac 1{n_2+\cfrac 1{n_3+\cfrac1{\ddots}}}}. \end{gather*} Thus we can express quotients of real numbers as \textbf{continued fractions.}\dindex{continued fraction}\dindexsub{fraction}{continued ---} In particular, we have \begin{equation*} \radix 2=1+\cfrac 1{2+\cfrac 1 {2+\cfrac 1{2+\cfrac 1{\ddots}}}}, \end{equation*} although we cannot here say exactly what this \emph{means.} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Give a \emph{geometrical} argument for the incommensurability of the diagonal and side of a square. (One way to start is to let $ABCD$ be a square. Draw a circle with center $A$ and radius $AC$. Extend $AB$ to meet the circle at $E$; extend $BA$ to meet the circle at $F$. Then $FB:BC::BC:BE$.) \item The expression for $\radix 2$ as a continued fraction determines a sequence of rational numbers that approaches $\radix 2$ as a limit. Calculate a few terms of this sequence, and find a recursive definition of the sequence. \end{enumerate} \section{Parity}\label{sect:parity} Here I propose one possible approach to the so-called \textsl{Boolean connectives}\tindexsub{Boolean}{--- connective}\tindexsub{connective}{Boolean ---}, which will be defined in \S~\ref{sect:connectives}. I also give a warning about how \emph{not} to write a proof. Every integer has a \textbf{parity,}\dindex{parity} which is $0$ if the integer is even, and $1$ if it is odd. Let the parity of the integer $x$ be denoted by \begin{equation*} \parity x.\glossary{$\operatorname{p}(x)$} \end{equation*} Some basic facts about evenness and oddness can be expressed in terms of this: \begin{lemma}\label{lem:par2} The equation $\parity{x+2}=\parity x$ is an identity. \end{lemma} \begin{proof} If $a$ is even, then so is $a+2$, so each member of the equation is $0$. If $a$ is odd, then so is $a+2$, so each member of the equation is $1$. Hence the equation is satisfied by all integers. \end{proof} The taking of parities respects multiplication in the following sense: \begin{lemma}\label{lem:par.} The equation $\parity{xy}=\parity x\parity y$ is an identity. \end{lemma} Parity respects addition too, but in a more complicated sense: \begin{lemma}\label{lem:par+} The equation $\parity{x+y}=\parity{\parity x+\parity y}$ is an identity. \end{lemma} Finally, applying the parity-operation twice is the same as applying it once:\footnote{Therefore parity can be called \textbf{idempotent.}\dindex{idempotent}} \begin{lemma}\label{lem:par-id} The equation $\parity{\parity x}=\parity x$ is an identity. \end{lemma} I have introduced parity so as to be able to define two new operations on $\Z$ in the following way. \emph{By definition} of the operations $\odot$\glossary{$x\odot y$} and $\oplus$,\glossary{$x\oplus y$} the following equations are identities: \begin{align*} x\odot y&=\parity{xy},\\ x\oplus y&=\parity{x+y}. \end{align*} These symbols are not standard, and they will not be used beyond the next section. I define two more operations. The following are also identities, by definition:\glossary{$x\ominus y$}\glossary{$x\sqcup y$} %\label{defn:lnot} \begin{align}\label{eqn:lnot} \ominus x&=x\oplus 1,\\ x\sqcup y&=(x\odot y)\oplus(x\oplus y). \end{align} For $\sqcup$, an alternative (but equivalent) definition is possible: \begin{theorem}\label{thm:id} The equation \begin{equation}\label{eqn:lor} x\sqcup y=\ominus(\ominus x\odot\ominus y) \end{equation} is an identity. \end{theorem} There are two ways we can proceed. \begin{proof}[Proof 1.] We reduce everything to the ordinary arithmetic operations. By the definitions and Lemma \ref{lem:par+}, we have the following chain of identities: \begin{align*} x\sqcup y&=(x\odot y)\oplus(x\oplus y)\\ &=\parity{(x\odot y)+(x\oplus y)}\\ &=\parity{\parity{xy}+\parity{x+y}}\\ &=\parity{xy+x+y}. \end{align*} Similarly, \begin{align*} \ominus(\ominus x\odot\ominus y)&=((x\oplus 1)\odot(y\oplus 1))\oplus 1&&\\ &=\parity{\parity{\parity{x+1}\parity{y+1}}+1}&&\\ &=\parity{\parity{\parity{(x+1)(y+1)}}+1}&&\text{[by Lemma \ref{lem:par.}]}\\ &=\parity{\parity{(x+1)(y+1)}+1}&& \text{[by Lemma \ref{lem:par-id}]}\\ &=\parity{\parity{(x+1)(y+1)}+\parity 1}&& \text{[by definition of parity]}\\ &=\parity{(x+1)(y+1)+1}&&\text{[by Lemma \ref{lem:par+}]}\\ &=\parity{xy+x+y+2}&&\text{[by arithmetic]}\\ &=\parity{xy+x+y}&&\text{[by Lemma \ref{lem:par2}]}. \end{align*} Our computations show that $x\sqcup y$ and $\ominus(\ominus x\odot\ominus y)$ are equal to the same thing (namely $\parity{xy+x+y}$); so they are equal to each other. This completes one possible proof. \end{proof} Alternatively, we show that all that matters is the parities of $x$ and $y$. \begin{proof}[Proof 2.] By definition of $\oplus$ and by Lemma \ref{lem:par+}, we have \begin{equation*} \parity x\oplus\parity y=\parity{\parity x+\parity y}= \parity{x+y}= x\oplus y. \end{equation*} By definition of $\odot$ and by Lemmas~\ref{lem:par.} and~\ref{lem:par-id}, we have \begin{equation*} \parity x\odot\parity y=\parity{\parity x\parity y}= \parity{\parity{xy}}= \parity {xy}=x\odot y. \end{equation*} Therefore, to verify any identity involving only $\odot$ and $\oplus$ (and operations derived from them, like $\ominus$ and $\sqcup$), it suffices to replace each variable with its parity. More precisely, to verify \eqref{eqn:lor}, it is enough to check the four possibilities when $x$ and $y$ are chosen from the set $\{0,1\}$. We have the following computations: \begin{equation*} \begin{array}{c|c||c|c|c||c|c|c|c} x & y & x\odot y & x\oplus y & x\sqcup y & \ominus x & \ominus y & \ominus x\odot\ominus y & \ominus(\ominus x\odot\ominus y)\\ \hline 0&0&0&0&0&1&1&1&0\\ 1&0&0&1&1&0&1&0&1\\ 0&1&0&1&1&1&0&0&1\\ 1&1&1&0&1&0&0&0&1 \end{array} \end{equation*} %\begin{center} % \begin{tabular}{c|c|c|c|c|c|c|c|c} %x & y & x\odot y & x\oplus y & x\sqcup y & \ominus x & % \ominus y & \ominus x\odot\ominus y & \ominus(\ominus x\odot\ominus y)\\ % \hline %0&0&0&0&0&1&1&1&0\\ %1&0&0&1&1&0&1&0&1\\ %0&1&0&1&1&1&0&0&1\\ %1&1&1&0&1&0&0&0&1 % \end{tabular} %\end{center} The columns headed by the two members of~\eqref{eqn:lor} are identical, so this equation is an identity. \end{proof} \emph{Either} of the two proofs just offered should be sufficient to establish the theorem as true. Note well the \emph{format} of the first proof. The aim was to arrive at~\eqref{eqn:lor}. The proof did not \emph{begin} with this equation; it began with one of the \emph{members} of the equation, and it showed that this member was equal to a new term. Then the \emph{other} member of~\eqref{eqn:lor} was shown to be equal to the same term. To write the proof as follows would \emph{not} be good style: \begin{equation}\label{eqn:bad-proof} \begin{aligned} x\sqcup y&\overset?=\ominus(\ominus x\odot\ominus y),\\ (x\odot y)\oplus(x\oplus y)&\overset?=((x\oplus 1)\odot(y\oplus 1))\oplus 1,\\ \parity{(x\odot y)+(x\oplus y)}&\overset?= \parity{\parity{\parity{x+1}\parity{y+1}}+1},\quad\\ \dots &\overset?=\dots,\\ \parity{xy+x+y}&= \parity{xy+x+y}. \end{aligned} \end{equation} Do not write proofs this way! It does not show the connexion between consecutive lines. Equations~\eqref{eqn:bad-proof} don't tell the reader that, for example, $\ominus(\ominus x\odot\ominus y)= ((x\oplus 1)\odot(y\oplus 1))\oplus 1$. In fact, the equations tell us \emph{nothing} that can be assumed to be correct. Think of the following example: \begin{equation}\label{eqn:1=1} \begin{aligned} -1&\overset?=1\\ (-1)^2&\overset?=(1)^2\qquad\\ 1&=1. \end{aligned} \end{equation} It certainly does not show that $-1=1$. If you are \emph{searching} for a proof of \eqref{eqn:lor}, then you might possibly write something like Equations \eqref{eqn:bad-proof}. Then, after you have found a correct line of argument, you should rewrite your findings before presenting them to somebody else as a proof. The next chapter will make this point again with the notion of \textsl{formal proof}\tindexsub{formal}{--- proof}\tindexsub{proof}{formal ---}: What one writes down when \emph{looking} for a formal proof is generally a lot different from the formal proof itself. \subsection*{Exercises} \begin{enumerate}\renewcommand{\labelenumi}{\theenumi.} \item Prove Lemmas~\ref{lem:par.}, \ref{lem:par+} and~\ref{lem:par-id}. \item Explain why the equations~\eqref{eqn:1=1} do not constitute a valid proof of the equation~$-1=1$. \item Suppose $\rightsquigarrow$ is a new arithmetic operation defined on the set $\{0,1\}$ as follows: \begin{equation*} \begin{array}{c|c||c} x&y&x\rightsquigarrow y\\ \hline 0&0&1 \\ 1&0&1 \\ 0&1&0 \\ 1&1&1 \end{array} \end{equation*} Find an arithmetic term $t$ such that the equation $\parity t=\parity x\rightsquigarrow\parity y$ is an identity. \end{enumerate} \section{Boolean connectives}\label{sect:connectives} In memory of George Boole,\footnote{See Boole himself~\cite[III.12, \textbf{[47]}, p.~51]{Boole}.} let us refer to the set $\{0,1\}$ as $\B$. In the last section, I defined some operations that convert integers into elements of $\B$. Considering the elements of $\B$ as integers, I want to restrict those operations on $\Z$ so as to apply \emph{only} to elements of $\B$. In so doing, I change their names: \glossary{$P\land Q$}\glossary{$P\eor Q$} \glossary{$\lnot P$}\glossary{$P\lor Q$} \begin{center} \begin{tabular}{c|c|c|c|c} on $\Z$: & $\otimes$ & $\oplus$ & $\ominus$ & $\sqcup$ \\ \hline on $\B$: & $\land$ & $\eor$ & $\lnot$ & $\lor$ \end{tabular} \end{center} I shall not use the four operations $\otimes$, $\oplus$, $\ominus$ and $\sqcup$ anymore. Operations on $\B$ can be called \textbf{(Boolean) connectives.}\dindexsub{Boolean}{--- connective}\dindexsub{connective}{Boolean ---} Specific English names can be given as follows: \begin{compactenum}[1)] \item $\land$ is \textbf{conjunction;}\dindex{conjunction} \item $\lnot$ is \textbf{negation;}\dindex{negation} \item $\lor$ is \textbf{(inclusive) disjunction;}\dindex{inclusive disjunction}\dindexsub{disjunction}{inclusive ---} \item $\eor$ is \textbf{exclusive disjunction}% \dindex{exclusive disjunction}% \dindexsub{disjunction}{exclusive ---} or \textbf{(material) non-equivalence.}% \dindexsub{material}{--- non-equivalence}% \dindexsub{non-equivalence}{material ---} \end{compactenum} Since $\B$ is finite, the definitions of connectives can be given in tables like the table in the last section: \begin{align*} & \begin{array}{c|c||c||c||c} P & Q & P\land Q & P\lor Q & P\eor Q\\ \hline 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 1 & 1\\ 1 & 1 & 1 & 1 & 0 \end{array}&& \begin{array}{c||c} P & \lnot P\\ \hline 0 & 1 \\ 1 & 0 \end{array} \end{align*} It will be convenient to have two more connectives, namely: \begin{compactenum}[1)]\setcounter{enumi}4 \item\label{page:imp}\glossary{$P\lto Q$} \textbf{(material) implication}% \dindexsub{material}{--- implication}% \dindexsub{implication}{material ---} or the \textbf{conditional:}% \dindex{conditional} $\lto$; \item\glossary{$P\liff Q$} \textbf{(material) equivalence}% \dindexsub{material}{--- equivalence}% \dindexsub{equivalence}{material ---} or the \textbf{biconditional:}% \dindex{biconditional} $\liff$. \end{compactenum} Again the definitions can be given in a table: \begin{center} \begin{tabular}{c|c||c||c} $P$ & $Q$ & $P\lto Q$ & $P\liff Q$\\ \hline $0$ & $0$ & $1$ & $1$ \\ $1$ & $0$ & $0$ & $0$ \\ $0$ & $1$ & $1$ & $0$ \\ $1$ & $1$ & $1$ & $1$ \end{tabular} \end{center} Certain identities should be evident: For example, $P\eor Q$ seems to mean the same thing as $\lnot(P\liff Q)$. Here though, we shall \emph{not} put a sign of equality% \index{sign of equality}% \index{equality!sign of ---} between the two expressions. Rather, as will be discussed more fully in \S~\ref{equivalent}, we shall write \begin{equation}\label{eqn:first-sim} P\eor Q\sim \lnot(P\liff Q), \end{equation} using the \textbf{swung dash}\dindex{swung dash}\dindexsub{dash}{swung ---}\dindexsub{symbol}{swung dash} $\sim$ rather than the sign $=$ of equality. Why? First, by analogy with the definition of arithmetic terms in \S~\ref{algebra}, we define \textbf{Boolean terms}\dindexsub{Boolean}{--- term}\dindexsub{term}{Boolean ---}\label{Boolean-terms} recursively as follows. Boolean terms are certain strings containing (some of) the following symbols: \begin{compactenum}[1)] \item $\land$, $\lnot$, $\lor$, $\eor$, $\lto$, $\liff$ (or other connectives, should we choose to define them); \item the \textbf{constants}\dindex{constant} $0$ and $1$; \item \textbf{variables}\dindex{variable} from the list $P_0$, $P_1$, $P_2$, \dots; \item the parentheses $($ and $)$. \end{compactenum} Then the Boolean terms are determined by the following rules: \begin{compactenum} \item Variables and constants are Boolean terms; \item If $\sv F$ is a Boolean term, then so is $\lnot {\sv F}$; \item If ${\sv F}$ and ${\sv G}$ are Boolean terms, then so is $(\sv F*\sv G)$, where $*$ is one of the connectives $\land$, $\lor$, $\eor$, $\lto$, $\liff$. \end{compactenum} Note that the constants $0$ and $1$ can also be considered as Boolean connectives, since they give values (namely, themselves) in $\B$. We could now define \Eng{Boolean polynomials,} and we could make from them what me might call \Eng{Boolean polynomial equations;} these would be examples of so-called \Eng{Boolean formulas.} We shall \emph{not} use such expressions however, since our main interest will lie in Boolean terms \emph{as such.} To suggest this, we shall refer to Boolean terms mainly as \textbf{(propositional) formulas.}\dindexsub{proposition}{---al formula}\dindexsub{formula}{propositional ---} As with arithmetic terms, so with propositional formulas, we can establish a conventional order of operations so as to avoid writing too many parentheses. We can always leave out an outer pair of parentheses. Then: %In formulas whose only non-constant connectives are $\land$, $\lnot$, %$\lor$, $\lto$ and $\liff$: \begin{compactenum}[1)] \item $\lnot$ has priority over all other connectives; \item $\land$ and $\lor$ have priority over $\lto$, $\liff$, and $\eor$; \item in case of two instances of $\lto$, the one on the \emph{right} has priority---we shall use this convention, because propositional formulas like $(P_0\lto(P_1\lto P_2))$ are more common than $((P_0\lto P_1)\lto P_2)$; so it will be convenient to let $P_0\lto P_1\lto P_2$ stand for the \emph{former;} \item in case of two instances of $\land$ or of $\lor$ or of $\eor$, the one on the right has priority---we could just as well give priority to the one on the left; we just want to allow ourselves to let strings like $P_0\land P_1\land P_2$ denote Boolean terms. \end{compactenum} Also, instead of writing variables $P_k$, we may use $\sv P$, $\sv Q$ and $\sv R$ instead. Similarly, we may use letters like ${\sv F}$, ${\sv G}$ and ${\sv H}$ to stand for formulas. The symbols $P_0$, $P_1$, and so on are the variables that can appear officially in propositional formulas. The symbols $\sv P$, $\sv Q$ are \textbf{syntactical variables;}\dindexsub{syntactic}{---al variable}\dindexsub{variable}{syntactic ---}\label{note:syntactical} in the sense of \cite[\S~08]{MR18:631a} we use them to \emph{refer} to the variables in formulas. Likewise, ${\sv F}$ and so on are not literally formulas; we use them as syntactical variables for formulas. \begin{examples}\label{first-example} By the order of operations, \begin{compactenum}[1)] \item the propositional formula denoted by ${\sv P}\lto {\sv Q}\lor {\sv R}$ is $({\sv P}\lto({\sv Q}\lor {\sv R}))$; \item $\lnot {\sv P}\land {\sv Q}$ is $((\lnot {\sv P})\land {\sv Q})$; \item ${\sv P}\land {\sv Q}\lor {\sv R}$ is ambiguous; the writer must say whether $({\sv P}\land {\sv Q})\lor {\sv R}$ or ${\sv P}\land ({\sv Q}\lor {\sv R})$ is intended; \item ${\sv P}\land {\sv Q}\land {\sv R}$ is $({\sv P}\land({\sv Q}\land {\sv R}))$; \item ${\sv P}\land {\sv Q}\land {\sv R}\lor {\sv P}$ is ambiguous; \item ${\sv P}\lto {\sv Q}\lto {\sv R}$ is ${\sv P}\lto({\sv Q}\lto {\sv R})$; \item ${\sv P}\eor {\sv Q}\eor {\sv R}$ is $({\sv P}\eor({\sv Q}\eor {\sv R}))$; \item ${\sv P}\lto {\sv Q}\land {\sv R}\lto {\sv S}$ is $({\sv P}\lto (({\sv Q}\land {\sv R})\lto {\sv S}))$. \end{compactenum} \end{examples} A propositional formula like $0\lto 1$ can be called \textbf{closed,}\dindex{closed formula} because it has no variables. By definition of the connective $\lto$, this formula $0\lto 1$ has the \textbf{value}\dindex{value}~$1$. The formulas $0\lto 1$ and~$1$ are not equal \emph{as formulas}; but the former can be considered as a \textbf{name}\dindex{name} for the latter (considered as an element of $\B$). Propositional formulas are so defined that every \emph{closed} formula is the name of a \emph{unique} element of $\B$. We shall prove this in \S~\ref{sect:unique}; meanwhile, some applications are in the following exercises. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item By the order of operations, which propositional formulas, if any, are denoted by the following? \begin{enumerate} \renewcommand{\labelenumii}{(\theenumii)} \item ${\sv P}\land\lnot {\sv Q}\eor {\sv R}\lor {\sv P}$; \item ${\sv P}\lto {\sv Q}\eor {\sv R}$; \item $P_0\lto P_1\lto P_2\lto P_3$; \item $P_0\lto P_1\lto\dots\lto P_n$. \end{enumerate} \item The following closed formulas are names of which elements of $\B$? \begin{enumerate} \renewcommand{\labelenumii}{(\theenumii)} \item $1\lto 1\lto 1$, \item $1\lto 0\lto 1$, \item $(0\lto 1)\liff 1$, \item $\lnot(0\eor 1)\liff(0\liff 1)$, \item $\lnot\lnot\lnot 0$, \item $(1\lor 0)\land 0$, \item $1\lor (0\land 0)$. \end{enumerate} \end{enumerate} \section{Propositional formulas and language}\label{sect:p-formulas} In one sense of the word, a \emph{model}\index{model} is a representation or description of something that one wants to build or understand. Think of an architect's model,\index{architect} or an orrery\index{orrery} (a model of the solar system). In this sense, symbolic logic can be seen as a model of ordinary language. In propositional logic, the Boolean connectives represent words such as \Eng{and, but, or, if,} and \Eng{not,} some of which are traditionally called \textsl{conjunctions}\tindex{conjunction}\dindexsub{word}{conjunction} (\Tur{ba\u gla\c clar}). Our main interest here is how such words affect the truth of statements, especially statements in mathematics. \subsection*{Truth} Aristotle defines truth in the \emph{Metaphysics} (IV, vii, 1: 1011 b 26). A literal translation of his words\footnote{The words are in \cite{Aristotle-XVII}: \Gk{t`o m`en g`ar l'egein t`o >`on m`h e>'inai >`h t`o m`h >`on e>~inai ye~udos, t`o d`e t`o >`on e>~inai ka`i t`o m`h >`on m`h e>~inai >alhj'es} (\Tur{Varl\i\u g\i n var olmad\i\u g\i n\i{} veya varolmayan\i n var oldu\u gunu s\"oylemek yanl\i\c st\i r. Bunu kar\c s\i l\i k varl\i\u g\i n var oldu\u gunu, var olmayan\i n var olmad\i\u g\i n\i{} s\"oylemek do\u grudur}).} is: \begin{quote} To declare the being not to be, or the not being to be, is false;---the being to be, and the not being not to be, is true. \end{quote} Alternatively, `It is false to say that what is, is not, or what is not, is; it is true to say that what is, is, and what is not, is not.' I propose (inspired by Alfred Tarski \cite{Tarski-T&P}) to refine this definition as follows: Let $A$ be a statement. Then: \begin{quote} $A$ is true if $A$, and $A$ is false if not $A$. \end{quote} This is a \emph{definition}; implicitly then, $A$ is true \emph{only} if $A$, and $A$ is false only if not~$A$. The definition is obscure. It becomes slightly less cryptic in an example where we can use the typographical convention established in the Preface: \begin{verse} \Eng{Grass is green} is true if grass is green;\\ \Eng{Grass is green} is false if grass is not green. \end{verse} Note what happens when we translate this into Turkish: \begin{verse} {\c Cimen ye\c silse, \Eng{Grass is green} do\u grudur;\\ \c cimen ye\c sil de\u gilse, \Eng{Grass is green} yanl\i\c st\i r.} \end{verse} \subsection*{Compound statements} We can now analyse certain compound statements. Let $A$ and $B$ be statements. \subsubsection*{Conjunctions, disjunctions, and negations} The statement \Eng{$A$ and $B$} is true if and only if $A$ and $B$; hence \Eng{$A$ and $B$} is true if and only if $A$ is true and $B$ is true. Compare this with the observation that ${\sv P}\land {\sv Q}$ takes the value $1$ if and only if ${\sv P}$ takes the value $1$ and ${\sv Q}$ takes the value $1$. If $1$ represents truth, then the connective $\land$ represents the conjunction \Eng{and}. The proposition \Eng{$A$~and~$B$} and the propositional formula ${\sv F}\land {\sv G}$ can alike be called \textbf{conjunctions.}\dindex{conjunction} Note well, however, that the proposition \Eng{$A$~and~$B$} belongs to \emph{our} ordinary language, while the formula ${\sv F}\land {\sv G}$ belongs to propositional logic. Similarly, \Eng{$A$ or $B$} is true if and only if $A$ is true or $B$ is true. Also, ${\sv P}\lor {\sv Q}$ takes the value $1$ if and only if ${\sv P}$ takes the value $1$, or ${\sv Q}$ takes the value $1$. So the connective $\lor$ represents the conjunction \Eng{or}. The proposition \Eng{$A$ or $B$} and the propositional formula ${\sv F}\lor {\sv G}$ can alike be called \textbf{disjunctions.}\dindex{disjunction} More precisely, $\lor$ represents \Eng{or} in its \emph{inclusive} sense. The \emph{exclusive} sense of \Eng{or} is intended in a sentence like \Eng{You may have tea or coffee after your meal}, if this means that you are allowed to have tea, and you are allowed to have coffee, but you are not allowed to have both. The exclusive \Eng{or} is represented by the connective $\eor$. The sentence \Eng{Not-$A$} is true if and only if $A$ is false; and $\lnot {\sv P}$ takes the value $1$ if and only if ${\sv P}$ takes the value $0$. If now $0$ represents falsity, then $\lnot$ represents \Eng{not}. Both \Eng{Not-$A$} and $\lnot {\sv F}$ can be called \textbf{negations.}\dindex{negation} (In fact the negation of an English statement is almost never formed simply by the prefixing of the word \Eng{not}; the \Eng{not} goes inside, perhaps with some other changes.) Mathematics often involves ignoring certain distinctions. From the propositions $A$ and $B$, we can form several compound propositions: \begin{center} \Eng{$A$ and $B$}\\ \Eng{$A$, but $B$}\\ \Eng{$A$; $B$} \end{center} Each of these may have its own rhetorical coloration, but we shall take them all to have the same truth-value. We may use for any of them the abbreviation \begin{equation*} A\amp B.\glossary{$A\amp B$} \end{equation*} (Note the slight typographical distinction between $\&$ and $\land$.) The sentence $A\amp B$ here is not a propositional formula; it is just a proposition or sentence of ordinary language. \subsubsection*{Implications} We can form some more compounds, all having the same truth-value: \begin{center} \Eng{If $A$, then $B$}\\ \Eng{When $A$, then $B$}\\ \Eng{$A$ implies $B$}\\ \Eng{$B$ if $A$}\\ \Eng{$B$, provided $A$}\\ \Eng{$A$ only if $B$} \end{center} These can be called \textbf{implications}\dindex{implication} and \textbf{conditional}\dindex{conditional} statements. Each of them has the \textbf{antecedent}\dindex{antecedent} $A$ and the \textbf{consequent}\dindex{consequent} $B$. We shall understand the compounds to be true if $B$ is true or $A$ is false (or both); otherwise, the compounds are false. We may use the abbreviation\glossary{$A\implies B$} \begin{equation*} A\implies B. \end{equation*} The propositional formula ${\sv P}\lto {\sv Q}$ can be analysed similarly, and we can apply the same terminology. The formulation \Eng{$B$ if $A$} can be understood as emphasizing that $A$ is a \textbf{sufficient condition}% \dindex{sufficient condition}% \dindexsub{condition}{sufficient ---} for $B$. The formulation \Eng{$A$ only if $B$} emphasizes that $B$ is a \textbf{necessary condition}% \dindex{necessary condition}% \dindexsub{condition}{necessary ---} for $A$. In ordinary language, the sentence \Eng{If $A$, then $B$} suggests causation. \Eng{If you drop that {\.Iznik} vase, then it will break}---you will cause the vase to break by dropping it. In mathematics though, the sentence \Eng{If $A$, then $B$} means no more than \Eng{$B$ is true or $A$ is false.} This is why the connective $\lto$ is called \textbf{material implication;}% \dindexsub{material}{--- implication}% \dindexsub{implication}{material ---}% \footnote{See the discussions in Church \cite[\S~05, n.~89, pp.~37f.]{MR18:631a} and Tarski \cite[\S\S~8,~9]{Tarski-Intro}; but these sources do not discuss the origin of the terminology.} it is to be distinguished from \textbf{formal implication,}% \dindexsub{formal}{--- implication}% \dindexsub{implication}{formal ---} that is, the implication suggested by a sentence like \Eng{If $A$, then $B$} in ordinary language. I suggest the following mnemonic device. In Platonic philosophy, the \emph{form} of something has a higher level of reality than its \emph{matter.}\footnote{I suspect Descartes alludes to this distinction when he says in the third of the \emph{Meditations on First Philosophy} (p.~41) \begin{quote} That this idea contains this or that objctive reality rather than some other one results from the fact that the idea gets its objective [that is, material?] reality from a cause in which there is at least as much formal reality as there is objective reality contained in the idea. \cite{Descartes-Med} Ama bu idea belirli bir nesnel olgusall\i k kapsad\i\u g\i{} i\c cin, onu hi\c c ku\c skusuz en az\i ndan kendisinin kapsad\i\u g\i{} nesnel olgusall\i k denli bi\c cimsel olgusall\i k kapsayan bir nedenden t\"uretiyor olmal\i d\i r. \cite{Descartes-Med-Tur} \end{quote}} In the sentence about a vase, there is a \emph{formal} connexion between antecedent and consequent: they both refer to the same vase, for example. Such a connexion is missing in a sentence like \Eng{If water is wet, then Constantine founded Constantinople}; but we count the sentence as `materially' true if we accept the consequent as true. (In this case, it is irrelevant that the antecedent is true.)\footnote{Elsewhere in mathematics, the term \emph{formal} is used to denote what might be called a \emph{lower} level of reality than usual. An expression $a+b$ may be called a \emph{formal sum} if $a$ and $b$ cannot `really' be added, except to produce the expression $a+b$.} There is a saying in English, \Eng{If wishes were horses, then beggars would ride.} We cannot analyse this as a material implication, simply because the antecedent and consequent are not propositions. We can try to recast the sentence as, \Eng{If wishes are horses, then beggars ride.} Then we can argue that the sentence is true, simply because the antecedent is false: wishes are \emph{not} horses. This observation says nothing about the truth of the original saying. In some mathematical writing, one sees statements like \begin{equation*} A\implies B\implies C. \end{equation*} This should be understood as an abbreviation for \begin{equation*} (A\implies B)\amp (B\implies C). \end{equation*} This conjunction is \emph{not} the same statement as the implication \begin{equation*} A\implies(B\implies C), \end{equation*} even though we understand the formula ${\sv F}\lto {\sv G}\lto {\sv H}$ as an abbreviation for the formula ${\sv F}\lto ({\sv G}\lto {\sv H})$. In mathematics, we often have occasion to write sentences like \Eng{$A$ is true, and therefore $B$ is true,} or more simply, \Eng{$A$, therefore $B$.} Logically, the truth-value of the sentence is the same as the truth-value of $A\amp B$; so please resist the temptation to write the sentence as $A\implies B$, or as \begin{gather*} \phantom{\implies{}}A\\ \implies B. \end{gather*} Instead of an arrow, just use words, such as \Eng{therefore, hence, consequently,} or \Eng{as a result.} \subsubsection*{Equivalences} In ordinary language, we can write indifferently \begin{center} \Eng{$A$ if and only if $B$}\\ \Eng{$A$ just in case $B$} \end{center} These are \textbf{equivalences}\dindex{equivalence} and \textbf{biconditional}\dindex{biconditional} statements, and for them we can use the abbreviation\glossary{$A\iff B$} \begin{equation*} A\iff B. \end{equation*} The formula ${\sv P}\liff {\sv Q}$ has a similar analysis and description. In mathematical writing, one may see statements like \begin{equation*} A\iff B\iff C; \end{equation*} this should be understood an abbreviation for \begin{equation*} (A\iff B)\amp (B\iff C). \end{equation*} \subsection*{Reasoning with compounds} Some fundamental rules of reasoning can be abbreviated thus: \begin{gather}\label{eqn:MP-ord} A\amp (A \implies B)\implies B;\\ \label{eqn:not-imp} \Enot(A\implies B)\iff A\amp\Enot B. \end{gather} (We are using a convention like that established in \S~\ref{sect:connectives}: the expression $\amp$ has priority over $\implies$ and $\iff$.) The operations of \textbf{conversion}% \dindex{conversion} and \textbf{contraposition}% \dindex{contraposition}% \dindexsub{proof}{contraposition} can be performed on implications: \begin{compactenum}[1)] \item the \textbf{converse}\dindex{converse} of $A\implies B$ is $B\implies A$; \item the \textbf{contrapositive}\dindex{contrapositive} of $A\implies B$ is $\Enot B\implies \Enot A$. \end{compactenum} The contrapositive of an implication is true if and only if the original implication is true: \begin{equation*} (A\implies B)\iff(\Enot B\implies\Enot A). \end{equation*} This observation is of great value in the proving of mathematical propositions. In particular, it often allows for proofs that are superior in style to proofs by contradiction. The stylistic problem with a proof by contradiction is that it contains false or even meaningless statements. The proof may still be correct, but it is inelegant. For example, in the proof of the Russell Paradox (Theorem~\ref{thm:Russell}), since $\Russell$ turns out not to be a set, the expressions $\Russell\in\Russell$ and $\Russell\notin\Russell$ turn out to be meaningless. A proof that avoids this problem is the following. \begin{proof}[Alternative proof of the Russell Paradox.]\label{Rus-alt} Let $A$ be an arbitrary set. Then either $A\in A$, or $A\notin A$. If $A\in A$, then $A\notin\Russell$, so $A\neq\Russell$. If $A\notin A$, then $A\in\Russell$, so again $A\neq\Russell$. Thus $A\neq\Russell$. That is, no set is equal to $\Russell$; so $\Russell$ is not a set. \end{proof} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Find a true implication whose converse is true. \item Find a true implication whose converse is false. \item Recast all foregoing proofs-by-contradiction to avoid any contradictions. \end{enumerate} \section{Quantifiers}\label{sect:quantifiers} As the Boolean connectives are used to model\index{model} the conjunctions of ordinary language, so the symbols called \textsl{quantifiers}\tindex{quantifier} can be used to model certain so-called \textsl{determiners,} especially \Eng{all} and \Eng{some.} Quantifiers are a part of \textsl{predicate logic.}\tindexsub{predicate}{--- logic}\tindexsub{logic}{predicate ---} In a section of the `XVII. Meditation' of his \emph{Devotions upon Emergent Occasions} of 1624, the clergyman and so-called metaphysical poet John Donne uses the determiners \Eng{no, every,} and \Eng{any,} in addition to the indefinite article \Eng{a(n).} The Meditation begins as follows (and here I preserve Donne's original spelling and typography, as found in \cite[pp.~440f.]{Donne}): `\textsc{Perchance} hee for whom this \emph{Bell} tolls, may be so ill, as that he knowes not it tolls for him;' later, the Meditation continues: \begin{quote} No man is an \emph{Iland,} intire of it selfe; every man is a peece of the \emph{Continent,} a part of the \emph{maine;} if a \emph{Clod} bee washed away by the \emph{Sea, Europe} is the lesse, as well as if a \emph{Promontorie} were, as well as if a \emph{Mannor} of thy \emph{friends} or of \emph{thine owne} were; any mans \emph{death} diminishes \emph{me,} because I am involved in \emph{Mankinde;} And therefore never send to know for whom the \emph{bell} tolls; It tolls for \emph{thee.} \end{quote} This text contains the following three clauses (and now I modernize the spelling): \begin{quote} No man is an island.\\ Every man is a piece of the continent.\\ Any man's death diminishes me. \end{quote} The first clause is contradicted some 350 years later by a verse of a popular song by Simon and Garfunkel \cite{S&G}: \begin{quote} I am a rock, I am an island. \end{quote} Donne says that the proposition \Eng{I am an island} is false, no matter who says it: it is false that some man is an island. (I take Donne's man to be a \emph{human being,} male or female.) So we can abbreviate the first two of Donne's clauses above by: \begin{quote} Not-(some $x$ is an island) \&\ (every $x$ is a piece of the continent), \end{quote} where the variable $x$ is understood to range over humanity. We can \emph{expand} this to \begin{quote} Not-(there is some $x$ such that $x$ is an island) \&\ (for every $x$, $x$ is a piece of the continent). \end{quote} The reason for this expansion is that the predicate \Eng{[is] an island} might be denoted by $P$, and \Eng{[is] a piece of the continent} might be denoted by $Q$. For the phrase \Eng{there is some $x$ such that}, we write \begin{equation*} \exists x;\glossary{$\exists x$} \end{equation*} for the phrase \Eng{for all $x$}, we write \begin{equation*} \forall x.\glossary{$\forall x$} \end{equation*} Then Donne's two clauses can be written \begin{equation*} \lnot\Exists xPx\amp \Forall xQx. \end{equation*} The symbol $\exists$ is the \textbf{existential quantifier;}\dindexsub{existential}{--- quantifier}\dindexsub{quantifier}{existential ---} the symbol $\forall$ is the \textbf{universal quantifier.}\dindexsub{universal}{--- quantifier}\dindexsub{quantifier}{universal ---}\footnote{With Chiswell and Hodges \cite{MR2319486}, one may prefer to say that the compound symbols $\Exists x$ and $\Forall x$ are the quantifiers.} We have just seen that these correspond respectively to the determiners \Eng{some} and \Eng{every,} and $\lnot\exists$ corresponds to \Eng{no.} We shall discuss, by and~by, what $\lnot\forall$ corresponds to. Let $\universe$ be some universal set as in \S~\ref{sect:sets}, let $P$ be a predicate, and let $A$ be the resulting set $\{x\in \universe\colon Px\}$. We can form several equations and inequations whose members are $\emptyset$, $A$ and $\universe$; with quantifiers, we can describe them. \begin{compactenum} \item $\Forall xPx$ means $A=\universe$. \item $\Exists xPx$ means $A\neq\emptyset$. \item $\lnot\Exists xPx$ means $A=\emptyset$. \item $\lnot\Forall xPx$ means $A\neq\universe$. \end{compactenum} The set denoted by \begin{equation*} \{x\in\universe\colon \lnot Px\} \end{equation*} consists of those elements of $\universe$ that are \emph{not} in $A$: it is the set \begin{equation}\label{eqn:Ac} A\comp,\glossary{$A\comp$} \end{equation} called the \textbf{complement}\dindex{complement} of $A$ (in $\universe$). Then we can form more equations, inequations and propositions on the pattern of those above: \begin{compactenum} \item $\Forall x\lnot Px$ means $A\comp=\universe$. \item $\Exists x\lnot Px$ means $A\comp\neq\emptyset$. \item $\lnot\Exists x\lnot Px$ means $A\comp=\emptyset$. \item $\lnot\Forall x\lnot Px$ means $A\comp\neq\universe$. \end{compactenum} But we have, for example, \begin{gather*} A\comp=\universe\iff A=\emptyset;\\ A\comp\neq\emptyset\iff A\neq\universe. \end{gather*} Correspondingly, we also have \begin{gather}\label{eqn:notE} \lnot\Exists x Px\iff\Forall x\lnot Px;\\ \label{eqn:notA} \lnot\Forall x Px\iff\Exists x\lnot Px. \end{gather} These equivalences are valuable tools for understanding propositions written with quantifiers. \begin{example} In calculus, \index{calculus!infinitesimal ---}% \index{infinite!---simal calculus} a function $f$ on $\R$ is said to be \textsl{continuous}% \tindex{continuous function} at a real number $a$ if, for every positive real number $\epsilon$, there is a positive real number $\delta$ such that, for every real number $x$, if $\abs{x-a}<\delta$, then $\abs{f(x)-f(a)}<\epsilon$. In our new symbolism, we can write the definition as \begin{equation}\label{eqn:ed} \Forall{\epsilon}(\epsilon>0\implies\Exists {\delta}(\delta>0\amp \Forall x(\abs{x-a}<\delta\implies\abs{f(x)-f(a)}<\epsilon))). \end{equation} Some people abbreviate this proposition to \begin{equation*} \Forall{\epsilon>0}\Exists {\delta>0} \Forall x(\abs{x-a}<\delta\implies\abs{f(x)-f(a)}<\epsilon). \end{equation*} By~\eqref{eqn:notE} and~\eqref{eqn:notA} above, as well as~\eqref{eqn:not-imp} in \S~\ref{sect:p-formulas}, along with the proposition that, in $\R$, $x0\land\Forall{\delta}(\delta>0\implies \Exists x(\abs{x-a}<\delta\land\abs{f(x)-f(a)}\geq\epsilon))), \end{equation*} which some people write as \begin{equation*} \Exists{\epsilon>0}\Forall{\delta>0}\Exists x(\abs{x-a}<\delta\amp\abs{f(x)-f(a)}\geq\epsilon). \end{equation*} For a specific example, let $f$ be the function given by \begin{equation*} f(x)= \begin{cases} \sin\displaystyle\frac 1x,&\text{ if }x\neq0;\\ 0,&\text{ if }x=0; \end{cases} \end{equation*} and $a=0$. We can show that $f$ is not continuous at $a$ as follows. The function $x\mapsto\sin x$ is periodic, with period $2\pi$: that is, \begin{equation*} \Forall x \sin(x+2\pi)=\sin x. \end{equation*} Also, $\sin(\pi/2)=1$. Let $\epsilon=1/2$. Say $\delta>0$. There is some integer $n$ greater than $1/2\pi\delta$. Then $2n\pi+\pi/2>2n\pi>1/\delta$. Let $x=1/(2n\pi+\pi/2)$. Then $\abs{x-a}=x<\delta$, but $\abs{f(x)-f(a)}=\abs{f(x)}=\sin(2n\pi+\pi/2)=1 \geq\epsilon$. This proves that $f$ is not continuous at $0$. \end{example} In~\eqref{eqn:ed}, note that the expression \begin{equation}\label{eqn:ed-core} \abs{x-a}<\delta\implies\abs{f(x)-f(a)}<\epsilon \end{equation} can be understood as a predicate with \emph{three} subjects---let's call them \textbf{arguments:}\dindex{argument} $x$, $\delta$, and $\epsilon$. If we wanted to abbreviate~\eqref{eqn:ed-core}, we might write it as $Sx\delta\epsilon$. Then $S$ is a \textbf{ternary}\dindex{ternary} predicate. Ternary predicates are common in mathematics, at least implicitly; for example, the equations \begin{align*} x+y&=z,&xy&=z \end{align*} can be understood as featuring ternary predicates. The signs $=$ and $<$ are \textbf{binary}\dindex{binary} predicates. \textbf{Singulary}% \dindex{singulary} predicates---predicates that take a single argument---are uncommon in mathematics, though they are needed in general treatments such as ours. \subsection*{Quantifier elimination and introduction} Let us return to the general setting where $\universe$ is some set, $P$ is a singulary predicate, and $A=\{x\in\universe\colon Px\}$. In proofs, there are several moves we might make that involve introducing or eliminating quantifiers from known propositions. \begin{compactdesc} \item[$\forall$-elimination:] If we know $\Forall xPx$, and $b$ is some element of $\universe$, then $b\in A$, so we can conclude \begin{equation*} Pb. \end{equation*} \item[$\exists$-introduction:] If $c$ is an element of $\universe$ such that $Pc$, then $c\in A$, so $a\neq\emptyset$, and therefore \begin{equation*} \Exists xPx. \end{equation*} \item[$\forall$-introduction:] If $b$ is an \emph{arbitrary} element of $\universe$, and we can show $Pb$, then it must be the case that \begin{equation*} \Forall xPx. \end{equation*} Of course it is essential that $b$ be \textbf{arbitrary.}\tindex{arbitrary} This means, in the proof of $Pb$, nothing about $b$ can be used, except its membership in $\universe$. \item[$\exists$-elimination:] Suppose $\Exists xPx$. If, by assuming $Pb$ for some unknown element of $\universe$, we are able to prove a proposition $\sigma$ that says nothing about $b$, then we can conclude that $\sigma$ is true. \end{compactdesc} These rules are illustrated in the next subsection. \subsection*{Prenex forms} In compound propositions involving quantifiers, it may be desirable to move all of the quantifiers to the front, in order to better understand the complexity of the proposition, or simply to avoid confusion. The result is said to be in \textbf{prenex}\dindex{prenex} form. For example,~\eqref{eqn:ed} can be rewritten in prenex form as \begin{equation*} \Forall{\epsilon}\Exists {\delta} \Forall x(\epsilon>0\implies(\delta>0\land(\abs{x-a}<\delta\implies\abs{f(x)-f(a)}<\epsilon))). \end{equation*} This is a consequence of the following lemmas, where $\sigma$ is a statement and $P$ is a singulary predicate. \begin{lemma} $(\sigma\implies\Exists xPx)\iff\Exists x(\sigma\implies Px)$. \end{lemma} \begin{proof} \begin{compactdesc} \item[($\bm{\Rightarrow}$)] Suppose $\sigma\implies\Exists xPx$. We consider two cases. Suppose first $\sigma$ is true. Then so is $\Exists xPx$, and hence for some $a$ in $\universe$ we have $Pa$ and therefore $\sigma\implies Pa$. By $\exists$-introduction, we can conclude $\Exists x(\sigma\implies Px)$. On the other hand, if $\sigma$ is false, then for all $a$ in $\universe$, we have $\sigma\implies Pa$. Since in particular there is \emph{some} $a$ in $\universe$, again by $\exists$-introduction we can conclude $\Exists x(\sigma\implies Px)$. \item[($\bm{\Leftarrow}$)] Suppose $\Exists x(\sigma\implies Px)$. By $\exists$-elimination, $\sigma\implies Pa$ for some $a$ in $\universe$. If $\sigma$ is false, then $\sigma\implies\Exists xPx$ is true. If $\sigma$ is true, then so is $Pa$, and therefore $\Exists xPx$ is true; hence also $\sigma\implies\Exists xPx$ is true.\qedhere \end{compactdesc} \end{proof} \begin{lemma} $(\sigma\amp\Forall xPx)\iff\Forall x(\sigma\amp Px)$. \end{lemma} \begin{proof} \begin{compactdesc} \item[($\bm{\Rightarrow}$)] Say $\sigma\amp\Forall xPx$. Let $a$ be arbitrary. then $Pa$ (by $\forall$-elimination), so $\sigma\amp Pa$, hence $\Forall x(\sigma\amp Px)$ by $\forall$-introduction (since $a$ was arbitrary). \item[($\bm{\Leftarrow}$)] Say $\Forall x(\sigma\amp Px)$. Let $a$ be arbitrary. Then $\sigma\amp Pa$ (by $\forall$-elimination), so $\Forall xPx$ by $\forall$-introduction (since $a$ was arbitrary) and hence $\sigma\amp\Forall xPx$. \qedhere \end{compactdesc} \end{proof} \begin{lemma}\label{lem:imp-all} $(\sigma\implies\Forall xPx)\iff\Forall x(\sigma\implies Px)$. \end{lemma} Now, writing $Sx\delta\epsilon$ for~\eqref{eqn:ed-core}, we can rewrite~\eqref{eqn:ed} as \begin{gather*} \Forall{\epsilon}(\epsilon>0\implies\Exists{\delta}(\delta>0\amp\Forall xSx\delta\epsilon)),\\ \Forall{\epsilon}\Exists{\delta}(\epsilon>0\implies(\delta>0\amp\Forall x Sx\delta\epsilon)),\\ \Forall{\epsilon}\Exists{\delta}(\epsilon>0\implies\Forall x(\delta>0\amp Sx\delta\epsilon)),\\ \Forall{\epsilon}\Exists{\delta}\Forall x(\epsilon>0\implies(\delta>0\amp Sx\delta\epsilon)). \end{gather*} The various rules must be applied with sensitivity to variables: \begin{lemma} $(\Forall xPx\implies\Forall xQx)\iff\Forall y\Exists x(Px\implies Qy)$. \end{lemma} \begin{proof} The following are equivalent. \begin{align*} &\Forall xPx\implies\Forall xQx,&&\\ &\Forall x(\Forall xPx\implies Qx),&&\\ &\Forall xPx\implies Qa&&\text{ for arbitrary $a$,}\\ &\lnot Qa\implies\Exists x\lnot Px&&\text{ for arbitrary $a$,}\\ &\Exists x(\lnot Qa\implies\lnot Px)&&\text{ for arbitrary $a$,}\\ &\Exists x(Px\implies Qa)&&\text{ for arbitrary $a$,}\\ &\Forall y\Exists x(Px\implies Qy).&&\qedhere \end{align*} \end{proof} \subsection*{Models} The assertion that the Diophantine equation $x^2-y^2=(x+y)(x-y)$ is an identity is the proposition \begin{equation*} \Forall x\Forall y x^2-y^2=(x+y)(x-y), \end{equation*} where $x$ and $y$ are understood to range over $\Z$. To express this last qualification, we can write\glossary{$\models$} \begin{equation*} \Z\models\Forall x\Forall y x^2-y^2=(x+y)(x-y) \end{equation*} (a notation to be developed in \S~\ref{sect:1st}). The expression $\Z\models\sigma$ can be read as one of \begin{center} \Eng{$\sigma$ is true in $\Z$,}\\ \Eng{$\Z$ satisfies $\sigma$,}\\ \Eng{$\Z$ is a model of $\sigma$;} \end{center} here $\Z$ is the \emph{context} in which $\sigma$ is true (see \S~\ref{sect:language}). The symbol $\models$ can be called the \textbf{semantic turnstile:}% \dindexsub{semantic}{--- turnstile}% \dindexsub{turnstile}{semantic ---} \emph{semantic,} because it concerns the \emph{meaning} of propositions (rather than the form), and \emph{turnstile,} because that is roughly what it looks like: a gate with a horizontal bar that you can turn away if you are allowed to pass (as for example when leaving the METU library). The \textsl{syntactic turnstile}% \tindexsub{syntactic}{--- turnstile}% \tindexsub{turnstile}{syntactic ---} $\proves$ will be introduced in \S~\ref{sect:formal}. The notation $\Z\nmodels\sigma$ means $\sigma$ is false in $\Z$, that is, $\Z\models\lnot\sigma$. \begin{example} The sentence $\Forall x(x\neq0\implies\Exists yxy=1)$ is false in $\Z$, but true in $\Q$, that is, \begin{align*} \Z&\nmodels\Forall x(x\neq0\implies\Exists yxy=1),& \Q&\models\Forall x(x\neq0\implies\Exists yxy=1). \end{align*} Hence also $\Z\models\Exists x(x\neq0\land\Forall yxy\neq1)$; for example, $\Z\models(2\neq0\land\Forall y2y\neq1)$. \end{example} \subsection*{Ordinary language} Look again at the equations \begin{equation*} A=\universe,\quad A\neq\emptyset,\quad A=\emptyset,\quad A\neq\universe. \end{equation*} These can be verbalized respectively as \begin{compactenum}[1)] \item {every}thing is in $A$, \item {some}thing is in $A$, \item {no}thing is in $A$, \item {not every}thing is in $A$. \end{compactenum} The first three of these clauses are obtained from the clause \Eng{thing is in $A$} by adding, respectively, a universal, an existential, and a negative determiner. The last clause needs the addition of \Eng{not every}; alternatively, the clause could be written as \Eng{something is not in $A$}. Apparently, in English, there is not a one-word expression with the meaning of \Eng{not every} and \Eng{some\dots not}. Some people might write the last clause on the list as \Eng{Everything is not in $A$,} or \Eng{All things are not in $A$.} For example, there is a saying: \begin{center} All that glitters is not gold. \end{center} It is pretty clear that what is meant is that \emph{some} things that glitter are not gold: some shiny attractive things are not worth much. But the saying looks as if it could be written as \Eng{All that glitters fails to be gold.} This does not have the intended meaning, since gold itself does glitter. To avoid possible misunderstanding, it seems better to write \begin{center} Not all that glitters is gold, \end{center} with \Eng{not} moved to the beginning. Turkish avoids the ambiguities possible from a misplaced \Eng{not}. In the Antalya \emph{otogar,} I once bought a bag of bananas with the brand name Asal. The bag displayed the slogan \begin{center} Her muz Asal muz de\u gildir. \end{center} This should be translated as \Eng{Not every banana is a Prime banana}. According to our understanding, the sentence \Eng{Every banana is not a Prime banana} would be rendered in Turkish as \begin{center} Hi\c cbir muz Asal muz de\u gildir. \end{center} The words \Eng{a(n)} and \Eng{any} are ambiguous. If you say \Eng{A dog has three legs}, you probably mean the \Eng{a} existentially: there is a dog that has three legs. But if you say \Eng{A dog has four legs}, probably you are describing dogs in general: every dog has four legs. The sentence \Eng{Anybody can come} could be a general invitation to everybody, or it could express a worry over the possibility that somebody will come. Still, the word \Eng{any} seems useful in ordinary life. Again, Donne writes: \begin{center} Any man's death diminishes me. \end{center} Could he write, instead, \Eng{Every man's death diminishes me}? In a mathematical context, the \Eng{every} is preferable; but \Eng{every man's death} suggests the image of all people dying at once; \Eng{any man's death} takes the deaths one by one. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove Lemma~\ref{lem:imp-all}. \item Find (with proof) prenex forms for the following: \begin{enumerate} \item $\sigma\amp\Exists xPx$, \item $\Forall xPx\implies\sigma$, \item $\Exists xPx\implies\sigma$, \item $\Exists xPx\implies\Forall xQx$, \item $\Forall xPx\implies\Exists xQx$. \end{enumerate} \item Rewrite $\Forall x\Exists yRxy$ in a form that does not use $\exists$. \item Write the negation of $\Exists x(Px\implies\Forall yRxy)$ in prenex form. \item Write the following sentences $\sigma$ in symbolic form, with quantifiers, and in each case, determine whether $\str M\models\sigma$, where $\str M$ is $\N$, $\Z$, $\Q$, or $\R$: \begin{enumerate} %\renewcommand{\labelenumii}{\theenumii)} \item every number has a square root; \item every positive number has a square root; \item for all coefficients $b$ and $c$, the equation $x^2+bx+c=0$ has two distinct solutions, provided $b^2\neq 4c$; \item there is no least number; \item between any two distinct numbers, there is another number. \end{enumerate} \end{enumerate} %\input{chapter-propositional.tex} \chapter{Propositional logic}\label{ch:logic} \setcounter{section}{-1} \section{Truth-tables}\label{sect:tt} Propositional formulas were defined in \S~\ref{sect:connectives}. It was suggested there that every \emph{closed} propositional formula ${\sv F}$ has a \textbf{value.}\dindex{value} Let us denote this value by \begin{equation*} \named {\sv F};\glossary{$\widehat {\sv F}$} \end{equation*} it is an element of $\B$ and can be found in the following way. First note that ${\sv F}$ meets one of the following conditions: \begin{compactenum}[1)] \item ${\sv F}$ is a constant from $\B$ (that is, $0$ or $1$), or \item ${\sv F}$ is $\lnot {\sv G}$ for some closed formula ${\sv G}$, or \item ${\sv F}$ is $({\sv G}*{\sv H})$ for some closed formulas ${\sv G}$ and ${\sv H}$, where $*$ is one of the connectives $\land$, $\lor$, $\lto$, $\liff$, and $\eor$. \end{compactenum} Then we can find $\named {\sv F}$ by the following \textbf{recursive}\dindex{recursive} procedure: \begin{compactenum} \item If ${\sv F}$ is in $\B$, then $\named {\sv F}$ is ${\sv F}$ itself. \item If ${\sv F}$ is $\lnot {\sv G}$, then $\named {\sv F}$ is the value of $\lnot\named {\sv G}$ as determined by the table in~\S~\ref{sect:connectives}. \item If ${\sv F}$ is $({\sv G}*{\sv H})$, then $\named {\sv F}$ is the value of $\named {\sv G}*\named {\sv H}$ as determined by the tables in~\S~\ref{sect:connectives}. \end{compactenum} In the terminology introduced at the end of \S~\ref{sect:connectives}, ${\sv F}$ is a \textbf{name}\dindex{name} for $\named {\sv F}$. It is proved in the next section below that $\widehat {\sv F}$ is \emph{uniquely} determined by the procedure just given for finding it; we can then indeed call $\widehat {\sv F}$ the \textbf{value,} or more precisely the \textbf{truth-value,}% \dindexsub{truth}{---{}-value}% \dindexsub{value}{truth-{}---} of ${\sv F}$. If a formula is not closed, then it does not have a value in $\B$. However, any formula can be made into a closed formula by \textsl{substitution}% \tindex{substitution} of values for its variables. For each propositional formula $\sv F$, there is some $n$ in $\N$ such that, for each $k$ in $\N$, if the variable $P_k$ appears in $\sv F$, then $kuto}, meaning \Eng{the same}. Originally a tautology was a redundant expression, such as \Eng{cease and desist}.} (The semantic turnstile% \index{semantic!--- turnstile}% \index{turnstile!semantic ---} $\models$ was introduced in \S~\ref{sect:quantifiers}. To be consistent with the notation in that earlier section, we might write~\eqref{eqn:semantic} as $\B\models {\sv F}$; but the variables in propositional formulas will always range over $\B$.) If ${\sv F}\sim 0$, we call ${\sv F}$ a \textbf{contradiction.}% \dindex{contradiction} We say ${\sv F}$ is \textbf{satisfiable}% \dindex{satisfiable} if it is not a contradiction. If both ${\sv F}$ and $\lnot {\sv F}$ are satisfiable, then ${\sv F}$ is a \textbf{contingency.}% \dindex{contingency} Hence, in the truth-table for ${\sv F}$, if the column for ${\sv F}$ itself contains: \begin{compactenum}[1)] \item only $1$s, then ${\sv F}$ is a tautology; \item only $0$s, then ${\sv F}$ is a contradiction; \item at least one $1$, then ${\sv F}$ is satisfiable; \item at least one $1$, and at least one $0$, then ${\sv F}$ is a contingency. \end{compactenum} Also, the following statements mean the same thing: \begin{compactenum}[1)] \item ${\sv F}\sim {\sv G}$; \item $\models {\sv F}\liff {\sv G}$; \item $\lnot({\sv F}\liff {\sv G})$ is not satisfiable. \end{compactenum} Thus, in effect, a test for equivalence is a test for tautology, which is a test for satisfiability. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Test for the equivalence of the following pairs of formulas by the truth-table method: \begin{enumerate} \item ${\sv P}$ and ${\sv Q}\lto {\sv P}$; \item ${\sv P}$ and ${\sv Q}\lto({\sv P}\land {\sv Q})$; \item ${\sv P}\lto({\sv Q}\lto {\sv R})$ and ${\sv P}\lto {\sv Q}\lto({\sv P}\lto {\sv R})$. \end{enumerate} \item Give examples of tautologies, contradictions, and contingencies. \item\label{exer:simp} Prove Lemma \ref{lem:simp}. \item\label{exer:and+1} Establish the following equivalences: \begin{enumerate} \renewcommand{\labelenumii}{(\theenumii)} \item\label{exer:not=1+} $\lnot {\sv P}\sim 1\eor {\sv P}$; \item ${\sv P}\lor {\sv Q}\sim {\sv P}\eor {\sv Q}\eor {\sv P}\land {\sv Q}$; \item ${\sv P}\eor {\sv Q}\sim {\sv Q}\eor {\sv P}$; \item $({\sv P}\eor {\sv Q})\eor {\sv R}\sim {\sv P}\eor {\sv Q}\eor {\sv R}$; \item ${\sv P}\land({\sv Q}\eor {\sv R})\sim {\sv P}\land {\sv Q}\eor {\sv P}\land {\sv R}$; \item ${\sv P}\eor {\sv P}\sim 0$. \end{enumerate} \item Is there a formula ${\sv F}$ such that \begin{equation*} \models ({\sv F}\lto ({\sv P}\liff {\sv Q}))\land({\sv P}\lor({\sv Q}\lor {\sv F}))? \end{equation*} (One way to solve this problem is to write out a truth table for $({\sv R}\lto ({\sv P}\liff {\sv Q}))\land({\sv P}\lor({\sv Q}\lor {\sv R}))$, then try to write a truth-table for $\sv F$. An alternative is to use the next two sections to write the original formula in an equivalent form $(\sv G\lto\sv F)\land(\sv F\lto\sv H)$, then check whether $\models\sv G\lto\sv H$.) \end{enumerate} \section{Substitution and replacement} If ${\sv F}$ is a formula for which $(e_0,\dots,e_{n-1})$ is a truth-assignment, then the constant formula ${\sv F}(e_0,\dots,e_{n-1})$ is obtained by \textbf{substitution.}\dindex{substitution} In this substitution, it is not essential that each $e_i$ be in the set $\B$, that is, $\{0,1\}$; if $({\sv G}_0,\dots,{\sv G}_{n-1})$ is a list of $n$ formulas, then from ${\sv F}$ we can obtain the formula\glossary{${\sv F}({\sv G}_0,\dots,{\sv G}_{n-1})$} \begin{equation*} {\sv F}({\sv G}_0,\dots,{\sv G}_{n-1}) \end{equation*} by \emph{substitution} of ${\sv G}_j$ for each instance of $P_j$ in ${\sv F}$, for each $j$ less than $n$. Note that, if we are using the usual infix notation (see \S~\ref{sect:unique}), but have removed parentheses as allowed by our conventions, then the substitutions must be done with parentheses as necessary to ensure that each substituted formula becomes a \emph{sub-formula} of the new formula. \begin{example} Suppose ${\sv F}$ is $P_0\land (P_1\lto P_0)$, and ${\sv G}_0$ is $P_0\lto P_1$, and ${\sv G}_1$ is $P_1\lto(P_0\lor P_2)$. Then ${\sv F}({\sv G}_0,{\sv G}_1)$ is \begin{equation*} (P_0\lto P_1)\land((P_1\lto(P_0\lor P_2))\lto P_0\lto P_1). \end{equation*} rather than $P_0\lto P_1\land(P_1\lto(P_0\lor P_2)\lto P_0\lto P_1)$. \end{example} Substitution is \textbf{associative}\dindex{associative} in that, if we substitute some formulas ${\sv G}_i$ into ${\sv F}$, and then substitute some formulas ${\sv H}_j$ into the result, we get the same formula as if we substitute the ${\sv H}_j$ first into the ${\sv G}_i$, and then the results into ${\sv F}$. Likewise, if you put a book in a box, then put the box on a table, you get the same result as if you first put the box on the table before putting the book in the box. The formal statement is the following: \begin{lemma}[Associativity]% \label{lem:sub}% \dindexsub{associativity}{A--- Lemma}% \dindexsub{lemma}{Associativity L---} Suppose ${\sv F}$ is an $n$-ary formula, and \begin{equation*} ({\sv G}_0, \dots, {\sv G}_{n-1}) \end{equation*} is a list of $n$ formulas, each one of them being $\ell$-ary. Let ${\sv H}$ be the formula ${\sv F}({\sv G}_0,\dots,{\sv G}_{n-1})$. Then ${\sv H}$ is $\ell$-ary. Suppose $(\sv K_0, \dots, \sv K_{\ell-1})$ is a list of $\ell$ formulas. Then the formula \begin{equation*} {\sv H}(\sv K_0,\dots,\sv K_{\ell-1}) \end{equation*} is the formula \begin{equation*} {\sv F}({\sv G}_0(\sv K_0,\dots,\sv K_{\ell-1}),\dots,{\sv G}_{n-1}(\sv K_0,\dots,\sv K_{\ell-1})). \end{equation*} Finally, suppose $\tuple e$ is a truth-assignment for the ${\sv G}_j$. Then $\tuple e$ is a truth-assign\-ment for ${\sv H}$. If also \begin{equation*} \named {\sv G}_j(\tuple e)= f_j \end{equation*} for each $j$ in $\{0,\dots,n-1\}$, then $(f_0,\dots,f_{n-1})$ is a truth-assignment $\tuple f$ for ${\sv F}$, and \begin{equation*} \named {\sv H}(\tuple e)=\named {\sv F}(\tuple f). \end{equation*} \end{lemma} \begin{proof} I claim that the proposition is obvious,\footnote{However, Church \cite[\S~15, p.~97]{MR18:631a} proves a version of this lemma by induction.} in the sense that no written proof will make the truth of the proposition clearer than it already is to the reader who has understood the proposition. \end{proof} Is a truth-assignment for ${\sv F}({\sv G}_0,\dots,{\sv G}_{n-1})$ also a truth-assignment for the ${\sv G}_j$? It is, if all of the variables $P_0$, \dots, $P_{n-1}$ actually \emph{appear} in ${\sv F}$; otherwise it may not be: \begin{example} Suppose ${\sv F}$ is just $P_0$, \emph{considered} as a binary formula. Let ${\sv G}_i$ be $P_i$ when $i\in\{0,1\}$. Then ${\sv F}({\sv G}_0,{\sv G}_1)$ is $P_0$. Now, $(0)$ is a truth-assignment for the formula $P_0$; but $(0)$ is not long enough to be a truth-assignment for ${\sv G}_1$. \end{example} \begin{theorem}[Substitution] \label{thm:substitution}\dindexsub{substitution}{S--- Theorem}\dindexsub{theorem}{Substitution Th---} If \begin{equation*}{\sv F}(P_0,\dots,P_{n-1})\sim {\sv G}(P_0,\dots,P_{n-1}),\end{equation*} and $({\sv H}_0,\dots,{\sv H}_{n-1})$ is a list of $n$ formulas, then \begin{equation*}{\sv F}({\sv H}_0,\dots,{\sv H}_{n-1})\sim {\sv G}({\sv H}_0,\dots,{\sv H}_{n-1}).\end{equation*} \end{theorem} \begin{proof} Since ${\sv F}\sim {\sv G}$, we have \begin{equation}\label{eqn:sub} \named {\sv F}(\tuple e)=\named {\sv G}(\tuple e) \end{equation} for all truth-assignments $\tuple e$ for ${\sv F}$ and ${\sv G}$. Let ${\sv F}'$ be ${\sv F}({\sv H}_0,\dots,{\sv H}_{n-1})$, and let ${\sv G}'$ be ${\sv G}({\sv H}_0,\dots,{\sv H}_{n-1})$. Suppose $\tuple f$ is a truth-assignment for the ${\sv H}_j$, and let $\named {\sv H}_j(\tuple f\;)=e_j$. Then \begin{align*} \named{{\sv F}'}(\tuple f\;) &=\named {\sv F}(\tuple e)&&\text{[by Lemma \ref{lem:sub}]}\\ &=\named {\sv G}(\tuple e)&&\text{[by \eqref{eqn:sub}]}\\ &=\named{{\sv G}'}(\tuple f\;)&&\text{[by Lemma \ref{lem:sub}].} \end{align*} Therefore ${\sv F}'\sim {\sv G}'$. This completes the proof.\footnote{This is also Burris's proof \cite[\S~2.3, pp.~46f.]{Burris}, although Burris's use of the fact given in Lemma~\ref{lem:sub} is not entirely explicit.} \end{proof} \begin{corollary}\label{cor:substitution} A tautology remains a tautology when arbitrary formulas are substituted for the variables. \end{corollary} \begin{example}\label{example:subst} Since ${\sv P}\lor\lnot {\sv P}$ is a tautology, so is $({\sv P}\lto {\sv Q})\lor\lnot({\sv P}\lto {\sv Q})$. \end{example} In ordinary language, the words \Eng{substitution} and \Eng{replacement} are nearly synonyms, although there is a distinction. From the expression $abc$, we get $adc$ in a way that can be described in two ways: \begin{compactenum} \item by replacing $b$ \emph{with} $d$, or \item by substituting $d$ \emph{for} $b$. \end{compactenum} When doing logic, we shall make another important distinction. If ${\sv F}$ is a sub-formula of ${\sv G}$, then we may \textbf{replace}\dindex{replace} ${\sv F}$ with another formula ${\sv F}'$. Here, to replace ${\sv F}$ is to replace a particular \emph{occurrence} of ${\sv F}$ (since possibly ${\sv F}$ appears more than once as a sub-formula of ${\sv G}$). \begin{example} In ${\sv P}\lor\lnot {\sv P}$, replacing the second occurrence of ${\sv P}$ with ${\sv Q}$ yields ${\sv P}\lor\lnot {\sv Q}$. \end{example} \begin{theorem}[Replacement]\label{thm:replacement} \dindexsub{replace}{R---ment Theorem}\dindexsub{theorem}{Replacement Th---} Suppose ${\sv F}$ is a sub-formula of ${\sv G}$, and \begin{equation*}{\sv F}\sim {\sv F}'.\end{equation*} Let ${\sv G}'$ be the result of replacing ${\sv F}$ with ${\sv F}'$ in ${\sv G}$. Then \begin{equation*}{\sv G}\sim {\sv G}'.\end{equation*} \end{theorem} \begin{proof} Say ${\sv G}$ is $n$-ary. Let ${\sv H}(P_0,\dots,P_n)$ be the result of replacing ${\sv F}$ with $P_n$ in ${\sv G}$. Then ${\sv G}$ itself is the formula \begin{equation*} {\sv H}(P_0,\dots,P_{n-1},{\sv F}), \end{equation*} and ${\sv G}'$ is ${\sv H}(P_0,\dots,P_{n-1},{\sv F}')$. The remainder of the proof\footnote{Burris \cite[\S~2.4, pp.~48ff.]{Burris} gives an elaborate proof using induction; but I think the work is unnecessary, once one has Lemma~\ref{lem:sub}. Church's proof \cite[\S~15, p.~101]{MR18:631a} leaves details to the reader, but also involves induction. Moreover, Church's proof refers to the principle of unique readability, which Burris seems not to discuss.} is an exercise involving Lemma~\ref{lem:sub}. \end{proof} \begin{corollary}\label{cor:replacement} A tautology remains a tautology when a sub-formula is replaced with an equivalent sub-formula. \end{corollary} \begin{example} Since $\models({\sv P}\lto {\sv Q})\lor\lnot({\sv P}\lto {\sv Q})$ by Example \ref{example:subst}, and \begin{equation*}\lnot({\sv P}\lto {\sv Q})\sim {\sv P}\land\lnot {\sv Q},\end{equation*} we have $\models({\sv P}\lto {\sv Q})\lor({\sv P}\land\lnot {\sv Q})$. \end{example} The Substitution and Replacement Theorems work together in the following way. From known equivalences, Substitution lets us derive many more. By Replacement, we can use these equivalences to write given formulas in different (but equivalent) form. That, in short, is the method of simplification, to be developed in \S~\ref{simplify}. Our first example of the procedure will be in \S~\ref{sect:adequacy}. Meanwhile, in \S~\ref{normal}, we shall describe some formulas such that \emph{every} formula is equivalent to one of them. These equivalences can be established by the procedure just described, using the stock of equivalences presented in Lemma~\ref{lem:simp}. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item If ${\sv F}({\sv P})$ is ${\sv P}\lto {\sv P}\lto {\sv P}$, what is ${\sv F}({\sv F}({\sv P}))$, written with the fewest possible parentheses? \item Prove Corollary \ref{cor:substitution}. \item Complete the proof of the Replacement Theorem (\ref{thm:replacement}). \item Prove Corollary \ref{cor:replacement}. \end{enumerate} \section{Normal forms}\label{normal} We noted in \S~\ref{algebra} that different arithmetic \emph{terms} may represent the same \emph{polynomial.} Among those terms, there may be a preferred term, which might be called a \textsl{normal form}% \tindexsub{normal}{--- form of a polynomial} of the polynomial. \begin{example} The normal form of $(5+x^2-2x)(1+x)-(x-1+2x^2)(x^3+6)$ might be \begin{equation*} 11-x-13x^2+2x^3-x^4-2x^5, \end{equation*} since the latter term is usually easier to work with. \end{example} If we have the truth-table of a formula, then we can read off an equivalent formula in so-called \textsl{disjunctive normal form.}% \tindexsub{disjunctive}{--- normal form}% \tindexsub{normal}{disjunctive --- form} The general procedure is described immediately, then illustrated by Example~\ref{example:dnf}. Suppose we have the truth-table for a formula ${\sv F}(P_0,\dots,P_{n-1})$. Say there are $m$ rows in which the entry for ${\sv F}$ itself is $1$. Then $m\leq 2^n$. If we ignore the other rows (namely, those rows in which the entry for ${\sv F}$ is $0$), then what remains has the form \begin{equation*} \begin{array}{c|c|c|c||c} P_0 & P_1 & \dots & P_{n-1} & {\sv F} \\ \hline \rule{0pt}{3ex} e_0^0&e_1^0&\dots&e_{n-1}^0&1\\ \rule{0pt}{3ex} e_0^1&e_1^1&\dots&e_{n-1}^1&1\\ \rule{0pt}{3ex} \vdots & \vdots & & \vdots & \vdots\\ \rule{0pt}{3ex} e_0^{m-1}&e_1^{m-1}&\dots&e_{n-1}^{m-1}&1 \end{array} \end{equation*} where each $e_j^i$ is in $\B$. If $i2$ can also be found (this is an exercise). How might we show that a certain signature is \emph{not} adequate? Note that the signature $\{\land,\lnot\}$ is adequate, even though it contains no nullary connectives: the two constant Boolean polynomials are represented in $\{\land,\lnot\}$ by ${\sv P}\land\lnot {\sv P}$ and $\lnot({\sv P}\land\lnot {\sv P})$ respectively. \begin{theorem} The signature $\{\land,\eor\}$ is \emph{not} adequate. \end{theorem} \begin{proof} We shall show that no formula in $\{\land,\eor\}$ represents $1$. Now, if \begin{equation*} {\sv F}(P_0,P_1,P_2,\dots,P_n)\sim 1, \end{equation*} then ${\sv F}(P_0,P_0,P_0,\dots,P_0)\sim 1$ by the Substitution Theorem. Hence it is enough to show that no \emph{singulary} formula in $\{\land,\eor\}$ represents $1$. In $\{\land,\eor\}$, we can represent $0$ by ${\sv P}\eor {\sv P}$. We also have \begin{align*} 0\land 0&\sim 0,& 0\eor 0&\sim 0,\\ 0\land {\sv P}&\sim 0,& 0\eor {\sv P}&\sim {\sv P},\\ {\sv P}\land 0&\sim 0,& {\sv P}\eor 0&\sim {\sv P},\\ {\sv P}\land {\sv P}&\sim {\sv P},& {\sv P}\eor {\sv P}&\sim 0. \end{align*} By the Replacement Theorem, we can create no singulary formula in $\{\land,\eor\}$ that is \emph{not} equivalent to $0$ or a variable. \end{proof} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove Theorem \ref{thm:and-not-ad}. \item\label{exer:not-to-ad} Prove that $\{\lnot,\lto\}$ is adequate. \item Prove that $\lnot$ by itself is \emph{not} adequate. \item Prove the adequacy of the \textbf{Sheffer stroke,}\dindex{Sheffer stroke}\dindexsub{connective}{Sheffer stroke} namely the connective $|$ such that ${\sv P}\mathrel{|}{\sv Q}\sim\lnot~({\sv P}\land {\sv Q})$. \item Find an adequate ternary ($3$-ary) connective. (See \S~\ref{sect:unique}, Exercise~\ref{exer:tern}.) \end{enumerate} \section{Simplification}\label{simplify} In proving Theorem \ref{thm:ad}, we used a known equivalence, and the Theorems of Substitution and Replacement, to `simplify' a formula in the sense of eliminating instances of disjunction. In the same way, we can simplify any formula to disjunctive normal form. The procedure relies on Lemma~\ref{lem:simp}. Using this lemma, given any formula, we can: \begin{compactenum}[1)] \item eliminate instances of $\lto$, $\liff$, and $\eor$; \item eliminate multiple negations, and make sure that the only arguments of $\lnot$ are variables; \item eliminate conjunctions of disjunctions; \item eliminate redundancies; now the formula is a disjunction of conjunctions of variables and negated variables, so we can finally: \item add variables as necessary to obtain a disjunctive normal form. \end{compactenum} \begin{example} Suppose ${\sv F}$ is the formula $\lnot({\sv P}\lto {\sv Q})\lor {\sv Q}$. The reduction of ${\sv F}$ to disjunctive normal form can proceed as follows: \begin{align*} {\sv F}&\sim \lnot(\lnot {\sv P}\lor {\sv Q})\lor {\sv Q}& & \text{[def'n of $\lto$]} \\ &\sim (\lnot\lnot {\sv P}\land\lnot {\sv Q})\lor {\sv Q}& & \text{[De Morgan]}\\ &\sim ({\sv P}\land\lnot {\sv Q})\lor {\sv Q}& &\text{[double negation]}\\ &\sim ({\sv P}\land\lnot {\sv Q})\lor({\sv Q}\land {\sv P})\lor({\sv Q}\land\lnot {\sv P}) & &\text{[new variable]}\\ &\sim ({\sv P}\land\lnot {\sv Q})\lor({\sv P}\land {\sv Q})\lor(\lnot {\sv P}\land {\sv Q}) & &\text{[commutativity]} \end{align*} \end{example} There may be more than one way to proceed: \begin{example} Let ${\sv F}$ be $\lnot(\lnot {\sv P}\lto {\sv Q})\land({\sv Q}\lor\lnot {\sv P})$. Then \begin{align*} {\sv F}&\sim \lnot(\lnot\lnot {\sv P}\lor {\sv Q})\land({\sv Q}\lor\lnot {\sv P}) && \text{[def'n of $\lto$]}\\ &\sim \lnot({\sv P}\lor {\sv Q})\land({\sv Q}\lor\lnot {\sv P}) && \text{[double neg.]}\\ &\sim (\lnot {\sv P}\land \lnot {\sv Q})\land({\sv Q}\lor\lnot {\sv P}) && \text{[De Morgan]}\\ &\sim ((\lnot {\sv P}\land \lnot {\sv Q})\land {\sv Q})\lor((\lnot {\sv P}\land\lnot {\sv Q})\land\lnot {\sv P}) && \text{[dist.]}\\ &\sim (\lnot {\sv P}\land (\lnot {\sv Q}\land {\sv Q}))\lor(\lnot {\sv P}\land(\lnot {\sv P}\land\lnot {\sv Q})) && \text{[assoc.; comm.]}\\ &\sim (\lnot {\sv P}\land 0)\lor((\lnot {\sv P}\land\lnot {\sv P})\land\lnot {\sv Q}) && \text{[red.; assoc.]}\\ &\sim 0\lor(\lnot {\sv P}\land\lnot {\sv Q}) && \text{[red.]}\\ &\sim \lnot {\sv P}\land\lnot {\sv Q} && \text{[red.]} \end{align*} Alternatively, \begin{align*} {\sv F}&\sim \lnot({\sv P}\lor {\sv Q})\land({\sv Q}\lor\lnot {\sv P}) && \text{[def'n of $\lto$; double neg.]}\\ &\sim (\lnot {\sv P}\land \lnot {\sv Q})\land({\sv Q}\lor\lnot {\sv P}) && \text{[De Morgan]}\\ &\sim \lnot {\sv P}\land (\lnot {\sv Q}\land({\sv Q}\lor\lnot {\sv P})) && \text{[assoc.]}\\ &\sim \lnot {\sv P}\land ((\lnot {\sv Q}\land {\sv Q})\lor (\lnot {\sv Q}\land \lnot {\sv P})) && \text{[dist.]}\\ &\sim \lnot {\sv P}\land (0\lor(\lnot {\sv Q}\land\lnot {\sv P})) && \text{[red.]}\\ &\sim \lnot {\sv P}\land (\lnot {\sv Q}\land\lnot {\sv P}) && \text{[red.]}\\ &\sim (\lnot\sv P\land\lnot\sv P)\land\lnot\sv Q&&\text{[comm; assoc]}\\ &\sim \lnot\sv P\land\lnot\sv Q&&\text{[red.]} \end{align*} \end{example} \begin{lemma}[Absorption Laws]\label{lem:abs-laws}\dindexsub{law}{Absorption L---s}\dindex{Absorption Laws} \begin{equation*} {\sv P}\land({\sv P}\lor {\sv Q}) \sim {\sv P},\qquad {\sv P}\lor({\sv P}\land {\sv Q}) \sim {\sv P}. \end{equation*} \end{lemma} If two formulas ${\sv F}$ and ${\sv G}$ are equivalent, then we can use simplification to show this as follows. \begin{compactenum} \item Simplify ${\sv F}$ to a disjunctive normal form ${\sv F}'$. \item Simplify ${\sv G}$ to a disjunctive normal form ${\sv G}'$. \item Note that ${\sv F}'\sim {\sv G}'$. (They should be the same formula, except possibly in the order of the constituents.) \end{compactenum} However, it may be easier to simplify directly from one formula to the other, or to use \emph{conjunctive} normal forms. \begin{example} The formulas $P_0\lto P_1\lto P_2$ and $P_1\lto P_0\lto P_2$ are equivalent, because \begin{align*} P_0\lto P_1\lto P_2&\sim \lnot P_0\lor(P_1\lto P_2)&& \text{[def'n of $\lto$]}\\ &\sim \lnot P_0\lor\lnot P_1\lor P_2&& \text{[def'n of $\lto$]}\\ &\sim \lnot P_1\lor\lnot P_0\lor P_2&& \text{[comm.]}\\ &\sim \lnot P_1\lor(P_0\lto P_2)&& \text{[def'n of $\lto$]}\\ &\sim P_1\lto P_0\lto P_2.&& \text{[def'n of $\lto$]} \end{align*} (Associativity was used silently.) The reduction of each formula to disjunctive normal form would be tedious, since that normal form is \begin{multline*} (\lnot P_0\land\lnot P_1\land\lnot P_2)\lor ( P_0\land\lnot P_1\land\lnot P_2)\lor (\lnot P_0\land P_1\land\lnot P_2)\lor{}\\ {}\lor(\lnot P_0\land\lnot P_1\land P_2)\lor ( P_0\land\lnot P_1\land P_2)\lor (\lnot P_0\land P_1\land P_2)\lor ( P_0\land P_1\land P_2); \end{multline*} but the conjunctive normal form is just the formula $\lnot P_0\lor\lnot P_1\lor P_2$, found in the original simplification. \end{example} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Given a formula in normal form, how would you write down its truth-table? \item Prove the Absorption Laws (\ref{lem:abs-laws}) using simplification. \item Use simplification to prove the following equivalences: \begin{enumerate} \item $\lnot({\sv P}\land {\sv Q})\lor {\sv R}\sim {\sv P}\land {\sv Q}\lto {\sv R}$; \item $({\sv P}\lto {\sv Q})\land({\sv R}\lto {\sv Q})\land \lnot {\sv Q}\lto\lnot({\sv P}\lor {\sv R})\sim 1$; \item ${\sv P}\lto({\sv Q}\lto {\sv R})\sim {\sv P}\lto {\sv Q}\lto({\sv P}\lto {\sv R})$; \item $({\sv P}\lor {\sv R})\land({\sv Q}\lor\lnot {\sv R})\sim ({\sv P}\land \lnot {\sv R})\lor({\sv Q}\land {\sv R})$; \item $(\sv P_0\lor\sv P_1)\land({\sv Q}_0\lor {\sv Q}_1)\sim \bigvee_{i<2}\bigvee_{j<2}(\sv P_i\land {\sv Q}_j)$. \end{enumerate} \item\label{exer:simpdnf} For $(\lnot P_0\lto P_1)\land(\lnot P_1\lto P_0)\lto(\lnot P_0\lor\lnot P_1)$, find the disjunctive normal form using simplification. \item Use simplification to verify the equivalences listed in \S~\ref{equivalent}, Exercise~\ref{exer:and+1}. \item Use simplification to establish ${\sv P}\liff {\sv Q}\sim {\sv P}\eor\lnot {\sv Q}$. \end{enumerate} \section{Logical entailment}\label{sect:entailment} Simplification is a way to prove that two formulas are logically equivalent. There are other relations between formulas that we may want to prove. If ${\sv F}$ is an $n$-ary formula such that $\named {\sv F}(\tuple e)$ for all truth-assignments $\tuple e$, then as in \S~\ref{equivalent} we write \begin{equation*} \models {\sv F}. \end{equation*} Suppose $({\sv F}_0,\dots, {\sv F}_m)$ is a list of $m+1$ formulas, each of them $n$-ary, such that, for all $n$-ary truth-assignments $\tuple e$, if $\named {\sv F}_i(\tuple e)=1$ for each $i$ in $\{0,\dots,m-1\}$, then $\named {\sv F}_m(\tuple e)=1$. Then we say that ${\sv F}_m$ is a \textbf{logical consequence}\dindexsub{logic}{---al consequence}\dindexsub{consequence}{logical ---} of $\{{\sv F}_0,\dots,{\sv F}_{m-1}\}$, or $\{\sv F_0,\dots,\sv F_{m-1}\}$ \textbf{logically entails}\dindexsub{logic}{---ally entails}\dindexsub{entails}{logically ---} $\sv F_m$, and we write \begin{equation*} {\sv F}_0,\dots,{\sv F}_{m-1}\models {\sv F}_m; \end{equation*} if the set $\{{\sv F}_0,\dots,{\sv F}_{m-1}\}$ is denoted by $\Sigma$, then we can also write \begin{equation*} \Sigma\models {\sv F}_m. \end{equation*} Logical entailment can in principle be established by truth-tables. However, this method is practical only when the numbers of variables and formulas are low. \begin{examples} \begin{enumerate} \item $\sv P\models\sv P\lor\sv Q$ because the table \begin{equation*} \begin{array}{ccc} \sv P&\lor&\sv Q\\\hline 0&0&0\\ 1&1&0\\ 0&1&1\\ 1&1&1 \end{array} \end{equation*} shows $\sv P\lor\sv Q$ is true whenever $\sv P$ is true. (It is irrelevant that $\sv P\lor\sv Q$ can be true when $\sv P$ is false.) \item Similarly $\sv P,\sv Q\models\sv P\land\sv Q$. \item $\sv P\lor\sv Q\fcom\sv Q\lto\sv R\models\sv P\lor\sv R$ by consideration of the starred rows in the table: \begin{equation*} \begin{array}{ccc|c|c|cc} \sv P&\sv Q&\sv R&\sv P\lor\sv Q&\sv Q\lto\sv R&\sv P\lor\sv R& \\\cline{1-6} 0 &0 &0 &0 &1 &0 & \\ 1 &0 &0 &1 &1 &1 &*\\ 0 &1 &0 &1 &0 &0 & \\ 1 &1 &0 &1 &0 &1 & \\ 0 &0 &1 &0 &1 &1 & \\ 1 &0 &1 &1 &1 &1 &*\\ 0 &1 &1 &1 &1 &1 &*\\ 1 &1 &1 &1 &1 &1 &* \end{array} \end{equation*} \end{enumerate} \end{examples} There are alternative methods for establishing logical entailment. The following should be compared with Lemma~\ref{lem:LL'ad}. \begin{lemma} If $\Sigma\models\sv F$, and $\Sigma\included\Sigma'$, then $\Sigma'\models\sv F$. \end{lemma} \begin{proof} If $\Sigma\models\sv F$, and $\Sigma\included\Sigma'$, and $\tuple e$ is a truth-assignment under which every formula in $\Sigma'$ is true, then every formula in $\Sigma$ is true under $\tuple e$, so $\named{\sv F}(\tuple e)=1$. This means $\Sigma'\models\sv F$. \end{proof} Corresponding to Theorem~\ref{thm:substitution}, we have \begin{theorem}[Substitution]\label{thm:conseq-sub} \dindexsub{substitution}{S--- Theorem} \dindexsub{theorem}{Substitution Th---} If $({\sv F}_0,\dots,{\sv F}_m)$ is a list of $n$-ary formulas such that \begin{equation*} {\sv F}_0,\dots,{\sv F}_{m-1}\models {\sv F}_m, \end{equation*} and $({\sv G}_0,\dots,{\sv G}_{n-1})$ is a list of $n$ formulas, then \begin{equation*} \sv H_0,\dots,\sv H_{m-1}\models\sv H_m, \end{equation*} where $\sv H_i$ is ${\sv F}_i({\sv G}_0,\dots,{\sv G}_{n-1})$ for each $i$ in $\{0,\dots,m\}$. \end{theorem} \begin{proof} Say $\tuple e$ is a truth-assignment for the ${\sv G}_j$ such that $\named {\sv H}_i(\tuple e)=1$ when $i}[r] & ({}\models {\sv F}_0\lto {\sv F}_1\lto \dots \lto {\sv F}_m) \ar@{=>}[d]\\ ({\sv F}_0,\dots,{\sv F}_{m-1}\proves[D] {\sv F}_m) \ar@{=>}[u] & ({}\proves[D] {\sv F}_0\lto {\sv F}_1\lto \dots \lto {\sv F}_m) \ar@{=>}[l] } \end{equation*} Suppose ${\sv F}_0,\dots,{\sv F}_{m-1}\models {\sv F}_m$. Then for every truth-assignment $\tuple e$ for the ${\sv F}_i$, either $\named {\sv F}_m(\tuple e)=1$, or $\named {\sv F}_i(\tuple e)=0$ for some $i$ in $\{0,\dots,m-1\}$. If $\named {\sv F}_i(\tuple e)=0$ and $i0$, then by the Deduction Theorem we have \begin{align*} P_0'\dots,P_{n-2}'&\proves[L]P_{n-1}\lto {\sv G},& P_0'\dots,P_{n-2}'&\proves[L]\lnot P_{n-1}\lto {\sv G}, \end{align*} so $P_0'\dots,P_{n-2}'\proves[L]{\sv G}$ by Lemma~\ref{lem:several}~\eqref{item:two-cases}. Continuing, we find $\proves[L]{\sv G}$, so ${\sv F}_0,\dots,{\sv F}_{m-1}\proves[L]{\sv F}_m$. \end{proof} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove Corollary~\ref{cor:D}. \item Prove Theorem~\ref{thm:L-sound}. \item Prove Lemma \ref{lem:conv-ded}. \item Prove parts (\ref{item:other-way}), (\ref{item:other-contrap}), (\ref{item:imp}) and (\ref{item:two-cases}) of Lemma \ref{lem:several}. \item Supply the missing details in the proof of Theorem \ref{thm:eval-corr}. \end{enumerate} \section{Compactness}\label{sect:compactness} So far, we have dealt with only finitely many formulas at once. But suppose $\Ey$ is a possibly infinite set of formulas. If $\psys N$ is a proof-system, then the expression \begin{equation*} \Ey\proves[N]\sv F \end{equation*} has the same meaning as before. So does the expression \begin{equation}\label{eqn:Ey} \Ey\models {\sv F}, \end{equation} except that there may be no $n$ such that each formula in $\Ey\cup\{\sv F\}$ is $n$-ary. A \textbf{truth-assignment,}\dindexsub{truth}{---{}-assignment} simply, is a function from $\N$ to $\B$. Then $\Ey$ is \textbf{satisfied}\dindex{satisfied} by a truth-assignment $k\mapsto e_k$ if $\named{\sv G}(e_0,\dots,e_{n-1})=1$ for every $n$-ary formula $\sv G$ in $\Ey$, for every $n$ in $\N$; and $\sv F$ is \textbf{true}\dindex{true} in this assignment if $\named{\sv F}(e_0,\dots,e_{n-1})=1$ (assuming $\sv F$ is $n$-ary). Then~\eqref{eqn:Ey} holds, by definition, if $\sv F$ is true in every truth-assignment that satisfies $\Ey$. Hence the following are equivalent: \begin{compactenum}[1)] \item $\Ey$ does not logically entail $\sv F$; \item $\Ey\cup\{\lnot {\sv F}\}$ is \textbf{satisfiable}\dindex{satisfiable} (that is, satisfied by some truth-assignment; note that this definition is compatible with the one in \S \ref{equivalent}). \end{compactenum} \begin{theorem}[Compactness]\label{thm:compactness}% \dindex{Compactness Theorem}% \dindexsub{theorem}{Compactness Th---} If every finite subset of a set of formulas is satisfiable, then the whole set is satisfiable. \end{theorem} \begin{proof} Suppose $\Ey$ is an infinite set of formulas such that every finite subset of $\Ey$ is satisfiable. For any $n$ in $\N$, let $\Ey_n$ consist of the $n$-ary formulas in $\Ey$. Then $\Ey_n$ is finite, so it is satisfiable by assumption. In particular, $\Ey_n$ is satisfied by a certain truth-assignment \begin{equation*}(e_0^n,e_1^n,\dots,e_{n-1}^n).\end{equation*} For each $n$, let such an assignment be chosen. So for each pair $(i,n)$ of natural numbers, if $i'atomos} \Eng{uncuttable, not compound,} from \Gk{t'omos} \Eng{a slice.}} formulas,% \tindex{atomic formula}% \tindexsub{formula}{atomic ---} which consist of terms joined by the sign of equality% \index{equality!sign of ---}% \index{sign of equality} or by a predicate. Atomic formulas can be preceded by quantifiers\index{quantifier} (with variables) or combined by means of Boolean connectives; formulas in general are obtained in this way. New constants symbolizing particular elements of $A$ can be used as \textsl{parameters}\tindex{parameter} in terms and formulas. \begin{example} The set $\Z$ of integers can be understood as a structure in the signature $\{0,1,-,+,\cdot,<\}$ (see \S~\ref{algebra}~\eqref{eqn:Z<}); a term in this signature (with parameters from $\Z$ as desired) is an \textsl{arithmetic term}\index{arithmetic!--- term}\index{term!arithmetic ---} as defined in \S~\ref{algebra}. Diophantine equations% \index{Diophantine equation}% \index{equation!Diophantine ---} and arithmetic inequalities% \index{arithmetic!--- inequality}% \index{inequality!arithmetic ---} are the {atomic} formulas in this signature. \end{example} The terminology of first-order logic is a means to give a precise but general account of some ideas that one encounters in high-school mathematics. \subsection*{Structures} By formal definition, a \textbf{structure}% %\dindex{structure} is an ordered pair $(A,\interpretation)$---which can also be referred to as $\str A$---where: \begin{compactenum}[1)] \item $A$ is a non-empty set, which is called the \textbf{universe}\dindex{universe} of the structure; \item $\interpretation$ is a function, written also as \begin{equation*} s\longmapsto s^{\str A}, \end{equation*} whose domain $\lang$ is called the \textbf{signature}\dindex{signature} of the structure; \item $s^{\str A}$ is either an element of $A$ or an $n$-ary operation or relation on $A$ for some positive integer $n$, for each $s$ in $\lang$. \end{compactenum} Here $\str A$ may be called a structure \textbf{of} $\lang$, or an \textbf{$\lang$-structure.}% \dindex{L-structure@$\lang$-structure}% \dindexsub{structure}{L-structure@$\lang$-structure} If $\lang=\{s_0,s_1,\dots\}$, then $\str A$ can be written as \begin{equation*} (A,s_0{}^{\str A}, s_{1}{}^{\str A},\dots), \end{equation*} or just as $(A,s_0, s_{1},\dots)$ unless ambiguity would result (that is, unless another structure of interest has the same universe and signature as $\str A$). Moreover, if the intended signature is clear, then $\str A$ may be written simply as $A$; that is, the universe may stand for the structure. The function $\interpretation$ is almost never referred to, except in general accounts like this one. \begin{examples}\label{examples:structures} The following are structures: \begin{compactenum}[1)] \item % $(\vnn,{}',0)$; $(\N,{}\scr{},0)$, or more briefly $\N$ (see \S~\ref{sect:sets}); \item the {power-set} structure\index{power!---{}-set structure} \index{structure!power-set ---}on a non-empty set $\Omega$, namely \begin{equation*} (\pow{\Omega},\emptyset,\Omega,\cap,\cup,{}\comp,\included); \end{equation*} \item the \textbf{truth-structure}% \footnote{This is not a standard term.}% \dindexsub{truth}{---{}-structure}% \dindexsub{structure}{truth-{}---} \begin{equation*} (\B,0,1,\land,\lor,\lnot{},\models), \end{equation*} where $\models$ is the binary relation $\{(0,0), (0,1), (1,1)\}$ on $\B$. \end{compactenum} \end{examples} The last two examples are the same if the elements of $\B$ are von-Neumann natural numbers% \index{von Neumann!--- natural number} and $\Omega$ is the von-Neumann natural number $1$. {Propositional logic}% \index{proposition!---al logic}% \index{logic!propositional ---} studies the truth-structure. The area of mathematics and logic called \textbf{model-theory}% \dindexsub{model}{---{}-theory}% \dindexsub{theory}{model-{}---} studies \emph{all} structures. When $\interpretation$ is as above in the structure $(A,\interpretation)$, and $s$ is an element of $\lang$, then: \begin{compactenum}[1)] \item $s^{\str A}$ is called the \textbf{interpretation}% \dindex{interpretation} in $\str A$ of $s$; \item $s$ is called a \textbf{symbol}\dindex{symbol} for $s^{\str A}$. \end{compactenum} So $s$ is one of the following, according to its interpretation: \begin{compactenum}[1)] \item a \textbf{constant;}\dindex{constant} \item an \textbf{$n$-ary function-symbol}% \dindexsub{function}{---{}-symbol}% \dindexsub{symbol}{function-{}---}% \dindexsub{nary@$n$-ary}{--- function-symbol} for some positive $n$ in $\vnn$; \item an \textbf{$n$-ary predicate}% \dindex{predicate}% \dindexsub{nary@$n$-ary}{--- predicate}% \dindexsub{symbol}{predicate} (or \textbf{relation-symbol}% \dindexsub{relation}{---{}-symbol}% \dindexsub{symbol}{relation-{}---}% \dindexsub{nary@$n$-ary}{--- relation-symbol}) for some positive $n$ in $\vnn$. \end{compactenum} Since nullary operations on $A$ can be considered as elements of $A$, a constant can be considered as a nullary function-symbol. Here are some observations about the definition of \Eng{structure}: \begin{asparaenum} \item I am following the old convention\footnote{Used for example by Chang and Keisler~\cite{MR0409165}. Recent writers (as Marker~\cite{MR1924282} and Rothmaler~\cite{MR1800596}) use `calligraphic' letters, not Fraktur: \begin{center} \begin{tabular}{r | c | c | c | c | c | c | c} For a structure with universe: & $A$ & $B$ & $C$ & \dots & $M$ & $N$ & \dots \\ \hline \rule{0mm}{4mm}I write: & $\str A$ & $\str B$ & $\str C$ & \dots & $\str M$ & $\str N$ & \dots\\ \hline \rule{0mm}{4mm}others may write: & $\mathcal A$ & $\mathcal B$ & $\mathcal C$ & \dots & $\mathcal M$ & $\mathcal N$ & \dots \end{tabular} \end{center} Another option, used by Hodges~\cite{MR94e:03002}, is to use an ordinary letter like $A$ for a structure, and then $\operatorname{dom}(A)$ for its universe. (Here \Eng{dom} stands for \textsl{domain.}% \tindex{domain})} of denoting the universe of a structure by a Roman letter, and the structure itself by the corresponding Fraktur or Gothic letter. One might not bother to make a typographical distinction between a structure and its universe. Indeed, as suggested in the examples, the distinction is not easy to make with standard structures like $\B$ or $\Z$ (which are commonly denoted by letters in a so-called blackboard-bold font). \item Similarly, it is not always easy or convenient to distinguish in writing between a symbol and its interpretation. \item In a structure $(A,\interpretation)$, the \textbf{interpretation-function}\dindexsub{interpretation}{---{}-function}\dindexsub{function}{interpretation-{}---} $\interpretation$ could be considered to carry, within itself, the universe $A$. In any case, $A$ and $\interpretation$ work together to provide interpretations of the symbols in $\lang$ \emph{as} elements of, or operations or relations on, a certain set, namely $A$ itself. That's all a structure is: something that provides a mathematical interpretation for certain symbols. What makes model-theory interesting is that the same symbols can have different interpretions. Here begins the distinction between \textbf{syntax}% \dindex{syntax}\label{syntax} (formal symbolism) and \textbf{semantics}% \dindexsub{semantic}{---s} (mathematical meaning).\footnote{The distinction was alluded to in \S~\ref{sect:quantifiers}. In propositional logic, formal entailment ($\proves$) can be understood as a syntactic notion, while logical entailment ($\models$) is semantic.} \end{asparaenum} \subsection*{Terms and formulas} The \textbf{terms}\dindex{term} of a first-order signature $\lang$ are conveniently written in Polish notation\index{Polish!--- notation}\index{notation!Polish ---} (see \S~\ref{sect:unique}). First, we introduce a list \begin{equation*} x_0,x_1, x_2,\dots \end{equation*} of \textbf{variables}\dindex{variable} (that is, \textbf{individual variables}\dindex{individual variable}\dindexsub{variable}{individual ---}: variables standing for \emph{individual} elements of a universe). Then, by definition, \begin{compactenum}[1)] \item all variables are terms of $\lang$; \item all constants of $\lang$ are terms of $\lang$; \item if $f$ is an $n$-ary function-symbol in $\lang$, and $(t_0,\dots,t_{n-1})$ is a list of $n$ terms of $\lang$, then \begin{equation*} ft_0\dotsb t_{n-1} \end{equation*} is a term of $\lang$; if $f$ is binary, then $ft_0t_1$ may also be written as \begin{equation*} (t_0\mathbin{f}t_1). \end{equation*} Finally, singulary function-symbols are sometimes written as superscripts on their arguments, as in $\scr n$ in \S~\ref{sect:sets}~\eqref{eqn:scrn} and $A\comp$ in \S~\ref{sect:quantifiers}~\eqref{eqn:Ac}. \end{compactenum} The \textbf{atomic formulas}% \dindex{atomic formula}% \dindexsub{formula}{atomic ---} are defined similarly: \begin{compactenum} \item If $t_0$ and $t_1$ are terms of $\lang$, then the equation \begin{equation*} t_0=t_1 \end{equation*} is an atomic formula of $\lang$; \item If $R$ is an $n$-ary predicate of $\lang$, and $(t_0,\dots,t_{n-1})$ is a list of $n$ terms of $\lang$, then the string \begin{equation*} Rt_0\dotsb t_{n-1} \end{equation*} is a term of $\lang$; if $R$ is binary, then $Rt_0t_1$ may also be written as \begin{equation*} t_0\mathrel Rt_1. \end{equation*} \end{compactenum} Finally, \textbf{formulas}\dindex{formula} in general can be defined: \begin{compactenum} \item Atomic formulas of $\lang$ are formulas of $\lang$. \item If $\phi$ is a formula of $\lang$, then so is $\lnot\phi$. \item If $\phi$ and $\psi$ are formulas of $\lang$, then $(\phi\land\psi)$ is a formula of $\lang$. \item If $\phi$ is a formula of $\lang$, and $x$ is an individual variable, then $\Exists x\phi$ is a formula of $\lang$. \end{compactenum} These are the \textbf{first-order formulas}% \dindexsub{first}{---{}-order formula}% \dindexsub{order}{first-{}--- formula}% \dindexsub{formula}{first-order formula} in the signature $\lang$; they constitute the \textbf{first-order logic}% \dindexsub{first}{---{}-order logic}% \dindexsub{order}{first-{}--- logic}% \dindexsub{logic}{first-order ---} in that signature. We can use other connectives in addition to, or instead of, $\land$ and $\lto$. One will generally want to use an adequate signature for propositional logic, like $\{\lnot,\land\}$ (Theorem~\ref{thm:and-not-ad}) or $\{\lnot,\lto\}$ (by \S~\ref{sect:adequacy}, Exercise~\ref{exer:not-to-ad}). Once the criterion of adequacy is met, then using fewer symbols makes the ensuing definitions and proofs easier to write down. We can also use the quantifier $\forall$; but formulas using $\forall$ can be rewritten with $\exists$ alone by means of~\eqref{eqn:notE} and~\eqref{eqn:notA} in \S~\ref{sect:quantifiers}. It is standard to write a formula $\lnot(t_0=t_1)$ as $t_0\neq t_1$. In the definition of formula, if the last condition is removed, then what is defined is the \textbf{quantifier-free formulas}\dindexsub{formula}{quantifier-free ---}\dindexsub{quantifier}{---{}-free formula} of $\lang$. \subsection*{Interpretations of terms} A term $t$ can be called \textbf{$n$-ary} if the set of its variables is a subset of $\{x_k:k0$.) \item Every constant $c$ is an $n$-ary term, interpreted in $\str A$ as the constant $n$-ary operation $\tuple x\mapsto c^{\str A}$ on $A$. (If $n=0$, then this operation can be understood as the element $c^{\str A}$ of $A$.) \item If $(t_0,\dots,t_{k-1})$ is a list of $n$-ary terms, and $f$ is a $k$-ary function-symbol, then the term $ft_0\dotsb f_{k-1}$ is $n$-ary and, as such, is interpreted in $\str A$ as the $n$-ary operation \begin{equation*} \tuple x\longmapsto f^{\str A}(t_0{}^{\str A}(\tuple x),\dots,t_{k-1}{}^{\str A}(\tuple x)) \end{equation*} on $A$. (If $n=0$, the interpretation is just the element $f^{\str A}(t_0{}^{\str A},\dots,t_{k-1}{}^{\str A})$ of $A$.) \end{compactenum} \begin{example} In $\Z$, the ternary terms $(x_0\cdot(x_1+x_2))$ and $((x_0\cdot x_1)+(x_0\cdot x_2))$ have the same interpretation, namely the ternary operation \begin{equation*} (x,y,z)\mapsto x(y+z) \end{equation*} on $\Z$. We could also write this operation more precisely as $(x,y,z)\mapsto x\cdot^{\Z}(y+^{\Z}z)$. (See \S~\ref{algebra} \eqref{eqn:identity}.) \end{example} \subsection*{Interpretations of formulas} Interpretations of formulas take longer to define precisely, but the idea is that $\lnot$, $\land$, and $\exists$ symbolize complementation, intersection, and \textsl{projection} respectively. An \emph{atomic} formula $\phi$ can be called \textbf{$n$-ary} if the set of its variables is a subset of $\{x_i\colon i}(0,0)(-6,-6)(6,6) \pscircle[linewidth=0.8mm](0,0)5 \end{pspicture} \hfill \begin{pspicture}(-6,-6)(6,6) \psaxes[labels=none]{->}(0,0)(-6,-6)(6,6) \psdots(5,0)(4,3)(3,4)(0,5)(-3,4)(-4,3)(-5,0)(-4,-3)(-3,-4)(0,-5)(3,-4)(4,-3) \end{pspicture} \hfill \begin{pspicture}(-6,-6)(6,6) \pscircle[linewidth=0.8mm,linestyle=dashed,fillstyle=solid,fillcolor=lightgray](0,0)5 \psaxes[labels=none]{->}(0,0)(-6,-6)(6,6) \end{pspicture} \hfill \begin{pspicture}(-6,-6)(6,6) %\pscircle[linestyle=dashed](0,0)5 \psaxes[labels=none]{->}(0,0)(-6,-6)(6,6) \psdots (2,-4)(1,-4)(0,-4)(-1,-4)(-2,-4) (3,-3)(2,-3)(1,-3)(0,-3)(-1,-3)(-2,-3)(-3,-3) (4,-2)(3,-2)(2,-2)(1,-2)(0,-2)(-1,-2)(-2,-2)(-3,-2)(-4,-2) (4,-1)(3,-1)(2,-1)(1,-1)(0,-1)(-1,-1)(-2,-1)(-3,-1)(-4,-1) (4,0)(3,0)(2,0)(1,0)(0,0)(-1,0)(-2,0)(-3,0)(-4,0) (4,1)(3,1)(2,1)(1,1)(0,1)(-1,1)(-2,1)(-3,1)(-4,1) (4,2)(3,2)(2,2)(1,2)(0,2)(-1,2)(-2,2)(-3,2)(-4,2) (3,3)(2,3)(1,3)(0,3)(-1,3)(-2,3)(-3,3) (2,4)(1,4)(0,4)(-1,4)(-2,4) \end{pspicture} \hfill\mbox{} \caption{Interpretations of $x_0{}^2+x_1{}^2=25$ and $x_0{}^2+x_1{}^2<25$.}\label{fig:circ-5} \end{figure} The interpretation in $\Z$ consists of the integer points on this circle, namely $(\pm 5,0)$, $(\pm 4,3)$, $(\pm 4,-3)$, $(\pm 3,4)$, $(\pm 3,-4)$, and $(0,\pm 5)$. The interpretation of $x_0{}^2+x_1{}^2<25$ in $\R$ is the interior of the disk bounded by the circle. \end{example} In the sense described in \S~\ref{sect:cartesian}, a nullary relation is a truth-value, $0$ or $1$. If $n=0$, then~\eqref{eqn:=int} and~\eqref{eqn:Rint} can be written as: \begin{gather}\label{eqn:=intE} (t_0=t_1)^{\str A}= \begin{cases} 1,&\text{ if } t_0{}^{\str A}=t_1{}^{\str A},\\ 0,&\text{ if }t_0{}^{\str A}\neq t_1{}^{\str A}; \end{cases}\\ \label{eqn:RintE} (Rt_0\dotsb t_{k-1})^{\str A}= \begin{cases}1,&\text{ if } (t_0{}^{\str A},\dots,t_{k-1}{}^{\str A})\in R^{\str A},\\ 0,&\text{ if } (t_0{}^{\str A},\dots,t_{k-1}{}^{\str A})\notin R^{\str A}. \end{cases} \end{gather} Quantifiers complicate matters, such as defining when a formula is $n$-ary. Assume that we \emph{have} defined this, and that $\phi$ and $\psi$ are arbitrary $n$-ary formulas, whose interpretations $\phi^{\str A}$ and $\psi^{\str A}$ are $n$-ary relations on $A$. Then the interpretations of $\lnot\phi$ and $(\phi\land\psi)$ are given by \begin{gather*} (\lnot\phi)^{\str A}=A^n\setminus\phi^{\str A}=(\phi^{\str A})\comp;\\ (\phi\land\psi)^{\str A}=\phi^{\str A}\cap\psi^{\str A}. \end{gather*} Now we have defined the interpretations of all \emph{quantifier-free} formulas. Suppose $\phi$ is an $(n+1)$-ary formula. Then $(\Exists{x_n}\phi)^{\str A}$ is an $n$-ary relation on $A$, namely the set of all $(a_0,\dots,a_{n-1})$ in $A^{n-1}$ such that $(a_0,\dots,a_{n-1},b)\in\phi^{\str A}$ for \emph{some} $b$ in $A$. This means \begin{equation}\label{eqn:n-1int} (\Exists{x_n}\phi)^{\str A}=\coordproj {n+1}n\setimb{\phi^{\str A}}, \end{equation} where $\coordproj{n+1}n$ is the function \begin{equation}\label{eqn:projection} (x_0,\dots,x_{n-1},x_n)\longmapsto(x_0,\dots,x_{n-1}) \end{equation} from $A^{n+1}$ to $A^n$; such a function can be called a \textbf{projection.}% \dindex{projection}% \dindexsub{function}{projection} (See Figure~\ref{fig:projection} and \S~\ref{sect:eq}.) \begin{figure}[t!] \begin{center} \begin{picture}(120,120)(-20,-20) \put(0,0){\line(0,1){100}} \put(0,0){\line(1,0){100}} \put(100,0){\line(0,1){100}} \put(0,100){\line(1,0){100}} \put(0,80){\circle*{5}} \put(-10,80){$b$} \put(30,80){\circle*{5}} \put(35,80){$(\tuple a,b)$} \thicklines \put(30,75){\vector(0,-1){70}} \put(30,0){\circle*{5}} \put(35,5){$\tuple a$} \put(35,40){$\coordproj{n+1}n$} \put(45,-20){$A^n$} \put(-20,45){$A$} \put(75,60){$A^{n+1}$} \end{picture} \end{center} \caption{Projection} \label{fig:projection} \end{figure} Note then that the formula $\Exists{x_n}\phi$ is considered as $n$-ary, not $(n+1)$-ary, even though it contains the variable $x_n$. The point is that this variable is not \textsl{free}% \dindex{free variable}% \dindexsub{variable}{free ---} in the formula; it is only \textsl{bound.}% \tindex{bound occurrence of variable} \begin{example*}[\ref{ex:25} continued] The formula $\Exists{x_1}x_0{}^2+x_1{}^2=25$ is singulary. Its interpretation in $\R$ is the interval $[-5,5]$; in $\Z$, the set $\{-5,-4,-3,0,3,4,5\}$. \end{example*} The set $\fv{\phi}$ of \textbf{free variables}% in a formula $\phi$ is defined recursively. \begin{compactenum} \item $\fv{\phi}$ is the set of variables appearing in $\phi$, if $\phi$ is atomic. \item $\fv{\lnot\phi}=\fv{\phi}$. \item $\fv{\phi\land\psi}=\fv{\phi}\cup\fv{\psi}$. \item $\fv{\Exists x\phi}=\fv{\phi}\setminus\{x\}$. \end{compactenum} Thus quantifiers \textbf{bind}\dindex{bind} variables, making them not free. \begin{example}\label{example:free} Suppose $R$ and $S$ are binary predicates. Then the free variables of \begin{equation*} \Exists x(x\mathrel Ry\land x\mathrel Sz) \end{equation*} are $y$ and $z$, but the free variables of \begin{equation*} \Exists xx\mathrel Ry\land x\mathrel Sz \end{equation*} are $x$, $y$, and $z$. \end{example} The second formula in the example is complicated by having bound \emph{occurrences}\index{occurrence} of $x$, even though $x$ is a free variable of the formula. In practice one avoids this situation by using instead a formula like $\Exists uu\mathrel Ry\land x\mathrel Sz$. For lack of a better term, let us refer to such a formula as \textbf{good.}% \dindexsub{good}{--- formula} There is a recursive definition of good formulas: \begin{compactenum} \item Atomic formulas of $\lang$ are good formulas of $\lang$. \item If $\phi$ is a good formula of $\lang$, then so is $\lnot\phi$. \item If $\phi$ and $\psi$ are good formulas of $\lang$, and every variable that occurs in both formulas is a \emph{free} variable of both formulas, then $(\phi\land\psi)$ is a good formula of $\lang$. \item If $\phi$ is a good formula of $\lang$, and $x$ is a free variable of $\phi$, then $\Exists x\phi$ is a good formula of $\lang$. \end{compactenum} \begin{example} If $\phi$ and $\psi$ are formulas, then so are $\Exists x\phi\land\Exists x\psi$ and $\Exists x\Exists x\phi$; but these are not \emph{good} formulas. \end{example} \begin{lemma}\label{lem:good} Suppose $x$ is a free variable of a good formula $\phi$, and $c$ is a constant. \begin{compactenum} \item In $\phi$, the variable $x$ never occurs right after $\exists$. \item The result of replacing each occurrence of $x$ in $\phi$ with $c$ is a good formula. \end{compactenum} \end{lemma} We never need work with any formulas other than good formulas. Also, restricting our attention to good formulas makes some general definitions easier. An arbitrary formula is \textbf{$n$-ary}\dindex{nary@$n$-ary}\dindexsub{arity}{nary@$n$-ary} if its \emph{free} variables are among $x_0$, \dots, $x_{n-1}$. If $\phi$ is such a formula, we may write it as \begin{equation*} \phi(x_0,\dots,x_{n-1}). \end{equation*} Suppose in particular $\phi$ is a \emph{good} formula. If $t_0$, \dots, $t_{n-1}$ are terms, we denote by \begin{equation*} \phi(t_0,\dots,t_{n-1}) \end{equation*} the formula that results from substituting $t_k$ for each occurrence of $x_k$ in $\phi$, \emph{provided} $x_k$ is actually a free variable of $\phi$. If $\phi$ is not necessarily good, then $\phi(t_0,\dots,t_{n-1})$ is the result of substituting $t_k$ for each free \textsl{occurrence}\tindex{occurrence} of $x_k$; but then one must define the free occurrences of variables. We avoid having to do this by restricting our attention to good formulas. \begin{example} If $\phi$ is ternary, and $\psi$ is $\Exists{x_1}\phi$, then $\psi$ is also ternary, but $\psi(c_0,c_1,c_2)$ is $\Exists{x_1}\phi(c_0,x_1,c_2)$. \end{example} The notation for substitution can be modified in an obvious way. If $\phi$ has at most one variable, $x$ (which could be $x_{1066}$, for all we know), then we can write $\phi$ as $\phi(x)$; then $\phi(t)$ is the result of substituting $t$ for $x$ in $\phi$, as long as $x$ really is a free variable of $\phi$; otherwise $\phi(t)$ is just $\phi$. A nullary formula is a \textbf{sentence.}\index{sentence} In a given signature $\lang$, a sentence $\sigma$ is either \textbf{true} or \textbf{false} in a structure $\str A$; if true, we write \begin{equation}\label{eqn:true-in} \str A\models\sigma; \end{equation} otherwise, $\str A\nmodels\sigma$. The definition is recursive: \begin{compactenum} \item If $\sigma$ is atomic, then \begin{equation*} \str A\models\sigma\iff\sigma^{\str A}=1. \end{equation*} \item If $\sigma$ is $\lnot\tau$, then \begin{equation*} \str A\models\sigma\iff\str A\nmodels\tau. \end{equation*} \item If $\sigma$ is $\tau\land\rho$, then \begin{equation*} \str A\models\sigma\iff\str A\models\tau\amp\str A\models\rho. \end{equation*} If $\sigma$ is $\Exists x\phi$, then $\str A\models\sigma$ if and only if \begin{equation*} \str A'\models\phi(c_b) \end{equation*} for \emph{some} element $b$ of $A$, where $c_b$ is a new constant, and $\str A'$ is the same as $\str A$, except that it interprets $c_b$ as $b$. \end{compactenum} Another way to write the last condition is as follows. Given the structure $\str A$ of $\lang$, we let $\lang(A)$ be $\lang$, together with a new constant $c_b$ for each element $b$ of $A$. Then we define $\str A_A$ as a structure of $\lang(A)$ in the obvious way: it interprets symbols of $\lang$ as $\str A$ does, and it interprets each new constant $c_b$ as $b$. If $\phi$ is a quantifier-free formula of $\lang(A)$, then we can denote by $\phi^{\str A}$ the interpretation of $\phi$ in $\str A_A$. If $\sigma$ is a sentence of $\lang(A)$, then we write $\str A\models\sigma$ if $\sigma$ is true in $\str A_A$. Usually we denote the constant $c_b$ by $b$. If $\sigma$ is $\Exists x\phi$, then we have simply that $\str A\models\sigma$ if and only if \begin{equation*} \str A\models\phi(b) \end{equation*} for some element $b$ of $A$. The following is an easy consequence of the definitions. \begin{lemma}\label{lem:qf} In some signature, if $\str A$ is a structure, and $\phi$ is a quantifier-free $n$-ary formula, then $\phi^{\str A}$ is the set of all $\tuple b$ in $A^n$ such that \begin{equation*} \str A\models\phi(\tuple b). \end{equation*} \end{lemma} Now we can use the lemma as a \emph{definition} of $\phi^{\str A}$ for arbitrary formulas $\phi$. Then the following is also easy. \begin{lemma} In some signature, if $\str A$ is a structure, and $\phi$ is a quantifier-free $(n+1)$-ary formula, then \begin{equation*} (\Exists{x_n}\phi)^{\str A}=\coordproj{n+1}n[\phi^{\str A}]. \end{equation*} \end{lemma} \subsection*{Entailment} Suppose $\sigma$ is a sentence of some signature $\lang$, and $\str A$ is a structure of $\lang$. If $\sigma$ is true in $\str A$, then we may say that $\str A$ is a \textbf{model}\dindex{model} of $\sigma$. More generally, if $\Gamma$ is a set of sentences of $\lang$, and each sentence in $\Gamma$ is true in $\str A$, then $\str A$ is a \textbf{model} of $\Gamma$; in this case, we may write \begin{equation}\label{eqn:model} \str A\models\Gamma. \end{equation} If $\sigma$ is true in \emph{every} model of $\Gamma$, then $\sigma$ is a \textbf{logical consequence}\dindexsub{logic}{---al consequence}\dindexsub{consequence}{logical ---} of $\Gamma$, or $\Gamma$ \textbf{logically entails}\dindexsub{logic}{---ally entails}\dindexsub{entails}{logically ---} $\sigma$, and we write \begin{equation}\label{eqn:entails} \Gamma\models\sigma. \end{equation} In case $\Gamma=\{\sigma_0,\dots,\sigma_{n-1}\}$, we may write also \begin{equation*} \sigma_0,\dots,\sigma_{n-1}\models\sigma. \end{equation*} If $n=0$ here, that is, $\Gamma$ is empty, then we write \begin{equation*} \models\sigma; \end{equation*} this means $\sigma$ is true in every structure of $\lang$, or in other words $\sigma$ is a \textbf{validity.}\index{validity} Note well that the semantic turnstile $\models$ has completely different meanings in~\eqref{eqn:model} and~\eqref{eqn:entails}. To avoid confusion, one might prefer to write~\eqref{eqn:model} as \begin{equation*} \models_{\str A}\Gamma. \end{equation*} Let us now permit $\lto$ in formulas, and define interpretations so that \begin{equation*} (\phi\lto\psi)^{\str A}=(\lnot(\phi\land\lnot\psi))^{\str A}. \end{equation*} Let us also permit $\forall$ in formulas, so that \begin{equation*} (\Forall x\phi)^{\str A}=(\lnot\Exists x\lnot\phi)^{\str A}. \end{equation*} A \textbf{generalization}\dindex{generalization} of a formula $\phi$ is a \emph{sentence} of the form $\Forall{u_0}\cdots\Forall{u_{n-1}}\phi$. A \textbf{tautology}\dindex{tautology} is a sentence $\sv F(\sigma_0,\dots,\sigma_{n-1})$, where $\sv F$ is a tautology of propositional logic. Let $\lang$ be a signature with infinitely many constants. We can now define a proof-system for $\lang$, in the sense of \S~\ref{sect:formal}, as follows. The only rule of inference is Detachment. The axioms are defined recursively: \begin{enumerate} \item Every tautology is an axiom. \item For all constants $c$ and $d$ and all singulary formulas $\phi$ of $\lang$, the following are axioms: \begin{align*} c&=c,& c=d&\lto\phi(c)\lto\phi(d). \end{align*} \item For all singulary formulas $\phi$ and $\psi$ of $\lang$ with free variable $x$, and all sentences~$\sigma$ of $\lang$, the following are axioms: \begin{gather*} \Forall x(\phi(x)\lto\psi(x))\lto\Forall x\phi(x)\lto\Forall x\psi(x),\\ \Forall x(\sigma\lto\psi(x))\lto\sigma\lto\Forall x\psi(x). \end{gather*} \item For all singulary formulas $\phi$ of $\lang$ with free variable $x$, the following is an axiom: \begin{equation*} \Exists x\phi(x)\lto\lnot\Forall x\lnot\phi(x). \end{equation*} \item If $\phi(x)$ is a formula of $\lang$ in which a constant $c$ of $\lang$ does not appear, and $\phi(c)$ is an axiom, then the following is an axiom: \begin{equation*} \Forall x\phi(x). \end{equation*} \end{enumerate} As in \S~\ref{sect:formal}, if $\Gamma$ is a set of sentences, and $\sigma$ is a sentence, we write \begin{equation}\label{eqn:sequent} \Gamma\proves\sigma \end{equation} if there is a formal proof of $\sigma$ from $\Gamma$ in the proof-system just defined. In this case, we may say that $\sigma$ is \textbf{deducible}\dindex{deducible} from $\Gamma$. The set of sentences deducible from a set $\Gamma$ is recursively defined: \begin{compactenum} \item It contains the axioms. \item It contains the sentences in $\Gamma$. \item If it contains $\sigma$ and $\sigma\lto\tau$, then it contains $\tau$. \end{compactenum} This allows the use of \textbf{induction}\dindex{induction} to prove statements about those sentences. Note that, by assuming that $\lang$ contains infinitely many constants, we ensures that, if $\Gamma$ does not formally entail $\sigma$ in $\lang$, then neither does it do so in a larger signature. \begin{theorem}[Soundness]\label{thm:1-soundness} If $\Gamma\proves\sigma$, then $\Gamma\models\sigma$. \end{theorem} \begin{proof} We use induction. The claim is trivially true when $\sigma\in\Gamma$. The claim is true when $\sigma$ is an axiom, since in that case $\models\sigma$ (exercise). Finally, suppose the claim is true when $\sigma$ is $\rho$ and when $\sigma$ is $\rho\lto\tau$. If these sentences are deducible from $\Gamma$, then by inductive hypothesis $\Gamma\models\rho$ and $\Gamma\models\rho\lto\tau$; therefore $\Gamma\models\tau$. \end{proof} The expression in~\eqref{eqn:sequent} can be called a \textbf{sequent.}\dindex{sequent} We usually do not write down formal proofs; we show that they exist by considering sequents. \begin{theorem}[Detachment] If $\Gamma\proves\rho$, and $\Gamma\proves\rho\lto\sigma$, then $\Gamma\proves\sigma$. \end{theorem} \begin{proof} If $\alpha_0,\dots,\alpha_m$ is a formal proof of $\rho$ from $\Gamma$, and $\beta_0,\dots,\beta_n$ is a formal proof of $\rho\lto\sigma$ from $\Gamma$, then \begin{equation*} \alpha_0,\dots,\alpha_m,\beta_0,\dots,\beta_n,\sigma \end{equation*} is a formal proof of $\sigma$ from $\Gamma$. \end{proof} \begin{theorem}[Deduction] If $\Gamma\cup\{\sigma\}\proves\tau$, then \begin{equation*} \Gamma\proves\sigma\lto\tau. \end{equation*} \end{theorem} \begin{proof} We use induction on $\tau$. There are three cases to consider. \begin{asparaenum} \item If $\tau$ is an axiom or an element of $\Gamma$, then \begin{align*} \Gamma&\proves\tau, &&\\ &\proves\tau\lto\sigma\lto\tau,&&\text{[tautology]}\\ \Gamma&\proves\sigma\lto\tau. &&\text{[Detachment]} \end{align*} \item If $\tau$ is $\sigma$, then $\sigma\lto\tau$ is a tautology, so again the claim follows. \item The last possibility is that $\rho$ and $\rho\lto\theta$ are deducible from $\Gamma\cup\{\sigma\}$, and the claim holds for each of these two sentences. Then \begin{align*} \Gamma&\proves\sigma\lto\rho,&&\text{[inductive hyp.]}\\ \Gamma&\proves\sigma\lto\rho\lto\theta,&&\text{[inductive hyp.]}\\ &\proves (\sigma\lto\rho) \lto(\sigma\lto\rho\lto\theta)\lto\sigma\lto\theta,&&\text{[tautology]}\\ \Gamma&\proves\sigma\lto\theta.&&\text{[Detachment (twice)]}\qedhere \end{align*}\qedhere \end{asparaenum} \end{proof} \begin{theorem}[Generalization] If $\Gamma\proves\phi(c)$, where $x$ is free in $\phi(x)$, and $c$ does not occur in $\phi(x)$ or in any sentence of $\Gamma$, then \begin{equation*} \Gamma\proves\Forall x\phi(x). \end{equation*} \end{theorem} \begin{proof} The claim is true when $\phi(c)$ is an axiom. The claim is vacuously true when $\phi(c)$ is in $\Gamma$, since then $c$ does occur in a sentence of $\Gamma$. The remaining possibility is that $\Gamma\proves\sigma$ and $\Gamma\proves\sigma\lto\phi(c)$. If $c$ does not occur in $\sigma$, then we may assume $\Gamma\proves\Forall x(\sigma\lto\phi(x))$. By Deduction from the appropriate axiom, $\Gamma\proves\Forall x\phi(x)$. The argument is nearly the same if $c$ does occur in $\sigma$. \end{proof} \begin{theorem}[Tautology] If in propositional logic, $\sv F_0,\dots,\sv F_{m-1}\models\sv G$, and in first-order logic, $\Gamma\proves\sv F_k(\sigma_0,\dots,\sigma_{n-1})$ when $k0\}$, because of the list: \begin{equation*} 1, \frac12,2, \frac13,3, \frac14,\frac23,\frac32,4, \frac15,5, \frac16,\frac25,\frac34,\frac43,\frac52,6, \frac17,\dots, \end{equation*} This list is made up of blocks of the form of \begin{equation*} \frac1n,\frac2{n-1},\frac3{n-2},\dots,\frac n1, \end{equation*} but with entries deleted if they are equal to entries that have already appeared. \item $\N\equip\Q$. \end{asparaenum} \end{examples} Thus there are sets $A$ and $B$ such that \begin{equation*} A\pincluded B\amp A\equip B. \end{equation*} But no such set $B$ can be \textsl{finite.} Indeed, to be precise, let us say that a set $C$ is \textbf{finite}\dindex{finite} if, for some $n$ in $\N$, \begin{equation*} C\equip\{0,\dots,n-1\}, \end{equation*} that is, for some $n$ in $\vnn$, \begin{equation*} C\equip n. \end{equation*} \begin{theorem}\label{thm:no-subset} No element $n$ of $\vnn$ has a proper subset $A$ such that $A\equip n$. \end{theorem} \begin{proof} We use induction. The claim is trivially true when $n=0$, since this has no proper subsets at all. Suppose the claim holds when $n=m$. Now let $n=m+1$, and suppose $A\subseteq n$, and $f$ is a bijection from $n$ to $A$. There are two cases to consider. \begin{compactenum} \item If $f(m)=m$, then $f\setminus\{(n,n)\}$ is a bijection from $m$ to $A\setminus\{m\}$, and the latter set is a subset of $m$. In this case, by inductive hypothesis, $A\setminus\{m\}=m$, so $A=n$. \item If $f(m)=k$, where $k0$, then the distinct elements of the quotient of $\Z$ by congruence \emph{modulo} $n$ are $[0]$, $[1]$, $[2]$, \dots, $[n-1]$. \item The function $[a,b]\mapsto a-b$ is a well-defined bijection from $\N^2\modsim$ to $\Z$. (In \S~\ref{sect:ZandQ}, the structure $\Z$ will be \emph{defined} in terms of $\N$ so that there is such a bijection.) \item \index{rational number}\index{number!rational ---}The function $[a,b]\mapsto a/b$ is a well-defined bijection from $\Z\times(\Z\setminus\{0\})/\mathord{\approx}$ to $\Q$. (In \S~\ref{sect:ZandQ}, the structure $\Q$ will be \emph{defined} in terms of $\Z$ so that there is such a bijection.) \item The equipollence-class of a set $A$ can be called the \textbf{cardinality}\dindex{cardinality} of $A$ and denoted by \begin{equation*} \size A. \end{equation*} Equipollent sets are sets having the same equipollence-class; such sets can also be said to have the same cardinality. An alternative definition of cardinality is given in \S~\ref{sect:cardinality}, whereby the cardinality of $A$ is a particular \emph{set} in the equipollence-class of $A$. \item The function $[\tuple x]\mapsto\coordproj nk(\tuple x)$ is a well-defined bijection from $A^n/\mathord{\sim_k^n}$ to $A^{n-1}$. \end{asparaenum} \end{examples} A \textbf{partition} of $A$ is a subset $P$ of $\pow A$ such that: \begin{compactenum}[1)] \item if $B$ and $C$ are in $P$, and $B\cap C\neq\emptyset$, then $B=C$; \item every element of $A$ is an element of some element of $P$. \end{compactenum} \begin{theorem}\label{thm:partition} If $\sim$ is an equivalence-relation on $A$, then $A\modsim$ is a partition of $A$. Conversely, if $P$ is a partition of $A$, then the relation \begin{equation*} \{(x,y)\in A^2:\Exsts XP\{x,y\}\subseteq X\} \end{equation*} is an equivalence-relation on $A$. \end{theorem} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove Theorem~\ref{thm:ff}. \item Prove Lemma~\ref{lem:[b]}. \item Prove Theorem~\ref{thm:well-defined}. \item Prove Theorem~\ref{thm:partition}. \item Let $A=\{0,1,2,3,4,5,6,7,8,9\}$. \begin{enumerate} \item Define an equivalence-relation $E$ on $A$ so that $\size{A/E}=5$. \item Can you define an equivalence-relation $F$ on $A$ so that $\size{A/F}=7$? \end{enumerate} \item Define an equivalence-relation $\sim$ on $\Z$ so that there is a bijection from $\Z\modsim$ to $\N$. \item For every property in the set $\{\text{reflexive, symmetric, transitive}\}$, find a set $A$ and a relation $R$ on $A$ that has just the other two properties. \item Suppose $R$ is a reflexive and symmetric relation on $A$, but $R\nincluded R/ R$. Can you find an equivalence-relation $S$ on $A$ such that $R\included S$, but $S\neq R\times R$? \end{enumerate} \section{Orderings} Let $R$ be a binary relation with field $A$. The following possible properties complement those given in \S~\ref{sect:eq}. The relation $R$ is: \begin{compactenum}[1)] \item \textbf{irreflexive,}\dindex{irreflexive}\dindexsub{reflexive}{ir---} if \begin{equation*} (A,R)\models\Forall x\lnot(x\mathrel Rx); \end{equation*} \item \textbf{anti-symmetric,} \dindex{anti-symmetric}\dindexsub{symmetric}{anti-{}---}if \begin{equation*} (A,R)\models\Forall x\Forall y(x\mathrel Ry\land y\mathrel Rx\lto x=y). \end{equation*} \end{compactenum} Again we have alternative characterizations. The relation $R$ is: \begin{compactenum}[1)] \item {irreflexive} if and only if $R\cap \Delta_A=\emptyset$; \item {anti-symmetric} if and only if $R\cap \conv R\included\Delta_A$. \end{compactenum} A reflexive, anti-symmetric, transitive relation is called a \textbf{partial ordering}% \dindexsub{partial}{--- ordering}% \dindexsub{order}{partial ---ing} of its domain. If $R$ is a partial ordering, and $A$ is its domain, then the structure $(A,R)$ is a \textbf{partially ordered set}% \dindexsub{partial}{---ly ordered set}% \dindexsub{order}{partially ---ed set}% \dindexsub{set}{partially ordered ---} or a \textbf{partial order.}% \dindexsub{partial}{--- order}% \dindexsub{order}{partial ---} More generally, we may say that a pair $(A,R)$ is a partial order when really it is $(A,R\cap A\times A)$ that is the partial order (see the examples below). A \textbf{strict partial ordering}% \dindexsub{strict}{--- partial ordering}% \dindexsub{partial}{strict --- ordering}% \dindexsub{order}{strict partial ---ing} is an irreflexive, anti-symmetric, transitive relation. If $R$ is a strict partial ordering, and the set $A$ \emph{includes} the domain of $R$, then the pair $(A,R)$ is a \textbf{strict partial order.}% \dindexsub{partial}{strict --- order}% \dindexsub{order}{strict partial ---}% \dindexsub{strict}{--- partial ordering} Note then that a strict partial order is technically \emph{not} a partial order (see Exercise~\ref{exercise:strict}). In any case, in the terminology used here, an \emph{order} is a kind of \emph{structure} (see Figure~\ref{fig:priene}); \begin{figure}[t!] \begin{center} \includegraphics[width=0.6\textwidth]{priene} \end{center} \caption[The temple at Priene: the Ionic order]{The remains of the temple at Priene: an example of the {Ionic order}\index{Ionic order}\index{order!Ionic ---} of architecture.\index{architect!---ure} Think of the columns as an order in our sense.} \label{fig:priene} \end{figure} an \emph{ordering} is the \emph{relation} that is part of an order. However, this terminological distinction is not of great importance. \begin{examples}\label{example:order} \mbox{} \begin{asparaenum} \item $(\pow A,\included)$ is a partial order; so is $(B,\included)$, if $B\included\pow A$. \item $(\pow A,\pincluded)$ is a strict partial order. \item (See the first of Examples~\ref{examples:equivalence2}.) We can understand logical entailment $\models$ as a binary relation on $\Fm n{\lang}/\mathord{\sim}$. Then $(\Fm n{\lang}/\mathord{\sim},\models)$ is a partial order. The case $n=2$ can be depicted as in Figure~\ref{fig:prop-as-order}. (Such a drawing of a partial order is called a \textsl{Hasse diagram.}\tindex{Hasse diagram}\tindexsub{diagram}{Hasse ---}) \begin{figure}[t!] \begin{equation*} %\xymatrix@R=0.45cm@C=0.1cm{ \xymatrix@!0{ &&&&&1&&&&\\ &&&&&&&&&\\ &\sv P\lor \sv Q \ar@{-}[uurrrr] &&& \sv P\lto \sv Q \ar@{-}[uur] && \sv Q\lto \sv P \ar@{-}[uul] && \sv P \shstroke \sv Q \ar@{-}[uulll] & \\ &&&&&&&&&\\ \sv Q \ar@{-}[uur] \ar@{-}[uurrrr] && \sv P \ar@{-}[uul] \ar@{-}[uurrrr] && \sv P\eor \sv Q\qquad \ar@{-}[uulll] \ar@{-}[uurrrr] & \qquad \sv P\liff \sv Q \ar@{-}[uul] \ar@{-}[uur] && \lnot \sv P \ar@{-}[uulll] \ar@{-}[uur] && \lnot \sv Q \ar@{-}[uulll] \ar@{-}[uul] \\ &&&&&&&&&\\ & \sv P\land \sv Q \ar@{-}[uul] \ar@{-}[uur] \ar@{-}[uurrrr] && \sv Q\nRightarrow \sv P \ar@{-}[uulll] \ar@{-}[uur] \ar@{-}[uurrrr] && \sv P\nRightarrow \sv Q \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uurrrr] &&& \sv P\curlywedge \sv Q \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uur] & \\ &&&&&&&&&\\ &&&&0 \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uur] \ar@{-}[uurrrr] &&&&& } \end{equation*} \caption[A partial order of propositional formulas]{In this depiction of the set of (truth-equivalence-classes of) propositional formulas in the variables $\sv P$ and $\sv Q$, $\sv F\models {\sv G}$ if and only if ${\sv G}$ can be reached from $\sv F$ by travelling upwards along the drawn lines. The new connective $\nRightarrow$\glossary{$P\nRightarrow Q$} here has the obvious meaning.} \label{fig:prop-as-order} \end{figure} \item $(\Z,{}\divides{})$ is a partial order. \item $(A,\Delta_A)$ is a partial order. \item $(A,\emptyset)$ is a strict partial order. \item The relation $\injects$ on sets is not a partial ordering; but we shall see in \S~\ref{sect:cardinality} that it `induces' a partial ordering of \emph{cardinalities}. \end{asparaenum} \end{examples} \begin{lemma}\mbox{} \begin{compactenum} \item If $(A,R)$ is a partial ordering, then $(A,R\setminus\Delta_A)$ is a strict partial ordering. \item If $(A,S)$ is a strict partial ordering, then $(A,S\cup\Delta_A)$ is a partial ordering. \end{compactenum} \end{lemma} In the lemma, one might say that $R\setminus\Delta_A$ is \textbf{associated}\dindex{associated} with $R$, and $S\cup\Delta_A$ with $S$. A partial order $(A,R)$ is a \textbf{linear order}% \dindex{linear order}% \dindexsub{order}{linear ---} or a \textbf{total order}% \dindex{total order}% \dindexsub{order}{total ---} if \begin{equation*} (A,R)\models\Forall x\Forall y(x\mathrel Ry\lor y\mathrel Rx), \end{equation*} that is, \begin{equation*} R\cup \conv R=A^2. \end{equation*} If\glossary{$\leq\quad<$} $\leq$ is a linear ordering, then the associated strict linear ordering can be denoted by $<$, and \emph{vice versa}. \begin{example} $(\Z,\leq)$ is a linear order; $(\Z,<)$ is a strict linear order. \end{example} Suppose $(A,R)$ and $(B,S)$ are partial orders, and $f\colon A\to B$. Then $f$ is \textbf{order-preserving}\dindexsub{order}{---{}-preserving} \dindexsub{preserve}{order-{}---ing}if \begin{equation*} a\mathrel R b\implies f(a)\mathrel S f(b) \end{equation*} for all $x$ and $y$ in $A$. An order-preserving function is an example of a more general notion: Suppose $\str A$ and $\str B$ are two structures in a signature $\lang$. A function $f$ from $A$ to $B$ is called a \textbf{homomorphism}\dindex{homomorphism}\dindexsub{function}{homomorphism} from $\str A$ to $\str B$ if \begin{equation}\label{eqn:homom} \str A\models\phi(a_0,\dots,a_{n-1})\implies \str B\models\phi(f(a_0),\dots,f(a_{n-1})) \end{equation} for all atomic formulas $\phi(x_0,\dots,x_{n-1})$ of $\lang$ and all $a_i$ in $A$, for all $n$ in $\N$. If~\eqref{eqn:homom} holds for all atomic and \emph{negated} atomic formulas $\phi(x_0,\dots,x_{n-1})$ of $\lang$ and all $a_i$ in $A$, for all $n$ in $\N$, then $f$ is an \textbf{embedding}% \dindex{embedding} of $\str A$ to $\str B$. Finally, $f$ is an \textbf{isomorphism}% \dindex{isomorphism, isomorphic structures}% \dindexsub{function}{isomorphism} if $f$ is invertible and $f\inv$ is a homomorphism from $\str B$ to $\str A$. A homomorphism is thus a function that \emph{preserves structure;} it \textbf{preserves}% \dindexsub{preserve}{---s} the symbols in a signature (hence it preserves the atomic formulas that use them). An embedding also preserves their complements; in particular, it preserves inequality, so it is an injection. The existence of an isomorphism shows that two structures are the \emph{same} as structures. If an isomorphism exists between $\str A$ and $\str B$, then $\str A$ and $\str B$ are called \textbf{isomorphic,} and we write \begin{equation*} \str A\cong\str B.\glossary{$\mathfrak A\cong\mathfrak B$} \end{equation*} Isomorphism is an equivalence-relation. \index{equivalence!---{}-relation} \index{relation!equivalence-{}---} Isomorphic structures have the same \emph{theories}\index{theory} (the proof is tedious, but not surprising). \begin{examples} \mbox{} \begin{asparaenum} \item An order-preserving function is a homomorphism of partial orders. An isomorphism of partial orders is an invertible order-preserving function whose inverse is also order-preserving. \item The identity is a homomorphism from $(\N,\divides)$ to $(\N,\leq)$, but not an embedding. \item Any function from a non-empty set to another is a homomorphism of sets. Equipollence is isomorphism of sets. \item By Theorem~\ref{thm:inv-homom}, if $f\colon A\to B$, then $X\mapsto f\inv\setimb X$ is a homomorphism from $(\pow B,\cap,\cup,{}\comp)$ to $(\pow A,\cap,\cup,{}\comp)$. \item More examples of homomorphisms and isomorphisms are in \S\S~\ref{sect:recursion},~\ref{sect:ZandQ} and~\ref{sect:reals}. \end{asparaenum} \end{examples} The following is a \textbf{representation theorem:}\dindexsub{represent}{---ation theorem}\dindexsub{theorem}{representation ---} it shows that every partial order \emph{can be represented by} (is isomorphic to) a structure of the form given in the first of the Examples~\ref{example:order}. Note how the proof of the theorem uses every property in the definition of partial orders. \begin{theorem} For every partial order $(A,R)$, there is a subset $B$ of $\pow{\Omega}$ such that $(A,R)\cong(B,\included)$. \end{theorem} \begin{proof} Let $f$ be the function from $A$ to $\pow A$ given by \begin{equation*} f(a)=\{y\in A:y\mathrel R a\}. \end{equation*} Then $f$ is injective: Indeed, suppose $c$ and $d$ are elements of $A$. If $c\mathrel Rd$ and $d\mathrel Rc$, then $c=d$ since $R$ is anti-symmetric. Suppose $c\neq d$. Then we may assume $\lnot(c\mathrel Rd)$. Then $c\notin f(d)$. But $c\in f(c)$ since $R$ is reflexive. Therefore $f(c)\neq f(d)$. Let $B=f\setimb A$; then $f$ gives a bijection between $A$ and $B$. Also, $f$ is order-preserving: Suppose $c\mathrel Rd$. If $e\in f(c)$, then $e\mathrel Rc$, so $e\mathrel Rd$ since $R$ is transitive; hence $e\in f(d)$. Thus $f(c)\included f(d)$. This shows that $f$ is order-preserving. But $X\mapsto f\inv\setimb X$ is also order-preserving (as a function on $B$, this set being equipped with the relation $\included$): If $f(c)\included f(d)$, then $c\in f(d)$ since $c\in f(c)$; so $c\mathrel Rd$. Therefore $f$ is an isomorphism from $(A,R)$ to $(B,\included)$. \end{proof} \begin{examples} \mbox{} \begin{asparaenum} \item The partial order $(\{1,2,3,4,5,6\},\divides)$ is isomorphic to $(B,\included)$, where $B$ is the set \begin{equation*} \{\{1\},\{1,2\}, \{1,3\}, \{1,2,4\}, \{5\}, \{1,2,3,6\}\}. \end{equation*} See Figure~\ref{fig:6}. \begin{figure}[t!] \begin{equation*} \xymatrix{ 4 & 6 & \\ 2 \ar@{-}[u] \ar@{-}[ur] & 3 \ar@{-}[u] & 5 \\ & 1 \ar@{-}[ul] \ar@{-}[u] \ar@{-}[ur] & }\qquad \xymatrix{ \{1,4\} & \{1,2,3,6\} & \\ \{1,2\} \ar@{-}[u] \ar@{-}[ur] & \{1,3\} \ar@{-}[u] & \{1,5\} \\ & \{1\} \ar@{-}[ul] \ar@{-}[u] \ar@{-}[ur] & } \end{equation*} \caption{Two isomorphic partial orders} \label{fig:6} \end{figure} \item A set of propositional formulas in $n$ variables, partially ordered by logical entailment $\models$, is isomorphic to a set of Boolean combinations of $n$ suitable sets, partially ordered by inclusion. Compare Figure~\ref{fig:prop-as-order} to Figure~\ref{fig:2sets}. \begin{figure}[t!] \begin{equation*} \xymatrix@!0{ &&&&&{\universe}&&&&\\ &&&&&&&&&\\ &A\cup B \ar@{-}[uurrrr] &&& A\comp\cup B \ar@{-}[uur] && A\cup B\comp \ar@{-}[uul] && A\comp\cup B\comp \ar@{-}[uulll] & \\ &&&&&&&&&\\ B \ar@{-}[uur] \ar@{-}[uurrrr] && A \ar@{-}[uul] \ar@{-}[uurrrr] && A\symdiff B \qquad \ar@{-}[uulll] \ar@{-}[uurrrr] & \qquad A\comp\symdiff B \ar@{-}[uul] \ar@{-}[uur] && A\comp \ar@{-}[uulll] \ar@{-}[uur] && B\comp \ar@{-}[uulll] \ar@{-}[uul] \\ &&&&&&&&&\\ & A\cap B \ar@{-}[uul] \ar@{-}[uur] \ar@{-}[uurrrr] && B\setminus A \ar@{-}[uulll] \ar@{-}[uur] \ar@{-}[uurrrr] && A\setminus B \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uurrrr] &&& A\comp\cap B\comp \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uur] & \\ &&&&&&&&&\\ &&&&{\emptyset} \ar@{-}[uulll] \ar@{-}[uul] \ar@{-}[uur] \ar@{-}[uurrrr] &&&&& } \end{equation*} \caption[A partial order of sets]{A partial order of sets. (The sets $A$ and $B$ here should be \emph{independent} in the sense that all Boolean combinations here are distinct.)} \label{fig:2sets} \end{figure} \end{asparaenum} \end{examples} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item\label{exercise:strict} Show that no partial ordering is a strict partial ordering. \item Are there partial orderings that are also equivalence-relations? \item Are there relations that are both symmetric and anti-symmetric? \item Write down the ordered pairs that belong to $\divides$, considered as a relation on $\{1,2,3,4,5,6\}$. Can you add pairs to this relation so that it becomes a linear ordering? \item More generally, if $R$ is a partial order on a finite set $A$, is there a linear ordering $S$ on $A$ such that $R\included S$? \item Find sets $A$ and $B$ such that all of the Boolean combinations depicted in Figure~\ref{fig:2sets} are distinct. \end{enumerate} \section{Infinitary Boolean operations} \label{sect:infinitary} The union of two sets is the set comprising everything that is in one or the other of the sets. There is no reason to restrict unions to two sets. Instead of writing $A\cup B$, we might write \begin{equation*} \bigcup\{A,B\}. \end{equation*} This is the \textsl{union} of the single set $\{A,B\}$, whose elements happen to be the sets $A$ and $B$. Then $\bigcup\{A,B,C\}$ is $A\cup B\cup C$, and so forth. If $\family S$ is a class of sets, then the \textbf{union}\dindex{union} of $\family S$ is the class \begin{equation*} \{x\colon\Exists y(y\in \family S\land x\in y)\}; \end{equation*} this is denoted by \begin{equation*} \bigcup \family S. \end{equation*} Unions in this new sense are \textbf{infinitary,}\dindexsub{infinitary}{--- union}\dindexsub{union}{infinitary ---} in the sense that the set $\family S$ may be infinite. We shall need the following for Theorem~\ref{thm:ord-sup}. \begin{axiom}[Union]\dindexsub{union}{U--- Axiom}\dindexsub{axiom}{A--- of Union} The union of a set is a set. \end{axiom} As there are infinitary unions, so there are \textbf{infinitary intersections:}\dindexsub{infinitary}{--- intersection}\dindexsub{intersection}{infinitary ---} If $\family S$ is a class of sets, then \begin{equation}\label{eqn:bigcap} \bigcap\family S=\{x\colon\Forall y(y\in \family S\lto x\in y)\}. \end{equation} So $A\cap B$ is $\bigcap\{A,B\}$, and so forth. \begin{theorem}\label{thm:int-set} The intersection of a non-empty set of sets is a set. \end{theorem} \begin{proof} If $\family S$ contains $A$, then \begin{equation*} \bigcap\family S=\{x\in A\colon\Forall y(y\in \family S\lto x\in y)\}, \end{equation*} which is a set by the Axiom of Separation, \ref{ax:separation}.\index{axiom!A--- of Separation}\index{separation!Axiom of S---} \end{proof} By strict application of~\eqref{eqn:bigcap}, $\bigcap\emptyset$ is the class of everything that belongs to some set; but this class is not a set. The following will be useful in the next chapter, starting in \S~\ref{sect:recursion}. \begin{theorem}\label{thm:in-A-un} Let $\family S$ be a set of sets, one of which is $A$. Then \begin{equation*} \bigcap \family S\included A\included \bigcup \family S. \end{equation*} \end{theorem} \begin{proof} Exercise. \end{proof} Sometimes, in an infinitary union $\bigcup\family S$ (or an intersection $\bigcap \family S$), the set $\family S$ is given as the range of a function. Say $f\colon A\to \pow B$. Then we can write \begin{equation*} \bigcap f\setimb A=\bigcap_{x\in A}f(x) \end{equation*} and $\bigcup f\setimb A=\bigcup_{x\in A}f(x)$. \begin{examples} \mbox{} \begin{enumerate} \item $\R=\bigcup_{n\in \N}(-1-n,n+1)$; \item $\bigcap_{n\in \N}[n,\infty)=\emptyset$; \item $\bigcap_{n\in \N}[-1/(n+1),1/(n+1)]=\{0\}$. \end{enumerate} \end{examples} \subsection*{Completeness} It is now possible to prove the completeness of our proof-system for first-order logic in a countable signature. \begin{theorem}[Completeness]\label{thm:1-completeness} Our proof-system for first-order logic in a countable signature is complete. \end{theorem} \begin{proof} Suppose $\Gamma$ is a set of sentences that does not formally entail $\sigma$. Then $\Gamma$ must be consistent (why?). We shall show that $\Gamma$ has a model. It will then follow that $\Gamma$ does not logically entail $\sigma$ (why?). Let $C$ be a set $\{c_k\colon k\in\N\}$ of new constants. We consider two cases. Suppose first that $\Gamma\cup\{c_i\neq c_j\colon i{\scr{}}>> \N\\ @V{h}VV @VV{h}V\\ A @>>{f}> A \end{CD} \end{equation*} That is, from the $\N$ on the left to the $B$ on the right, there are two different routes, but each one yields the same result. In fact, the theorem is simply that there is a unique \emph{homomorphism}\index{homomorphism} from $(\N,{}\scr{},0)$ to $(A,f,b)$. A \textbf{recursive definition,}\dindexsub{recursive}{--- definition}\dindexsub{definition}{recursive ---} or a \textbf{definition by recursion,}\dindexsub{definition}{--- by recursion}\dindexsub{recursion}{definition by ---} is a definition of a function on $\N$ that is justified by Theorem \ref{thm:recursion}. Informally, we can define such a function $h$ by specifying $h(0)$ and by specifying how $h(\scr n)$ is obtained from $h(n)$. Sections \ref{sect:arith-ops} and \ref{sect:recursion-gen} will provide several important examples of recursive definitions. Such definitions are sometimes\footnote{Dedekind calls them definitions by induction in \cite[Theorem 126, p.~85]{MR0159773}, which corresponds to the Recursion Theorem above.} called \textsl{inductive definitions,}\tindexsub{inductive}{--- definition}\tindexsub{definition}{inductive ---} or \textsl{definitions by induction}.\tindexsub{definition}{--- by induction}\tindexsub{induction}{definition by ---} However, this terminology is misleading when \axi{} is called the Axiom of Induction. Logically, the Recursion Theorem is equivalent to the three Peano Axioms together; the Recursion Theorem is strictly stronger than the Induction Axiom, in the sense that there are models of \axi{} that do not satisfy Theorem~\ref{thm:recursion}. The remainder of this section is devoted to proving this. Let us say that a structure \textbf{admits (definition by) recursion}\dindexsub{admits}{--- definition by recursion}\dindexsub{definition}{admits --- by recursion}\dindexsub{recursion}{admits definition by ---} if it satisfies the Recursion Theorem. That is, a structure $\str A$ in the signature $\{{}\scr{},0\}$ admits recursion if and only if, for any other structure $\str B$ in this signature, there is a unique homomorphism from $\str A$ to $\str B$. Similarly, structures that satisfy the Induction Axiom can be said to \textbf{admit (proof by) induction.}\dindexsub{admits}{--- proof by induction}\dindexsub{proof}{admits --- by induction}\dindexsub{induction}{admits proof by ---} \begin{theorem}\label{thm:isom} All structures that admit recursion are isomorphic. \end{theorem} \begin{proof} Suppose $\str A$ and $\str B$ admit recursion. Then there are unique homomorphisms $f$ from $\str A$ to $\str B$ and $g$ from $\str B$ to $\str A$. Hence the composition $g\circ f$ is a homomorphism from $\str A$ to itself; so it is the unique such homomorphism. But $\id_A$ is also such a homomorphism. Therefore $g\circ f=\id_A$. Similarly, $f\circ g=\id_B$. Therefore $g=f\inv$, by Theorem~\ref{thm:bij}. \end{proof} \begin{corollary}\label{cor:rec-Peano} All structures that admit recursion satisfy the Peano axioms; in particular, they admit induction. \end{corollary} \begin{proof} By the theorem, every structure that admits recursion is isomorphic to $(\N,{}\scr{},0)$. This satisfies the Peano axioms; hence so does every structure isomorphic to it. \end{proof} However, there are structures that admit induction, but not recursion:\footnote{Apparently Peano himself did not recognize the distinction between proof by induction and definition by recursion; see the discussion of Landau~\cite[p.~x]{MR12:397m}. Burris~\cite[p.~391]{Burris} does not acknowledge the distinction. Stoll~\cite[p.~72]{MR83e:04002} uses the term `definition by weak recursion', although he observes that the validity of such a definition does \emph{not obviously} follow from the Induction Axiom. However, Stoll does not \emph{prove} (as we have done in Example \ref{exam:ind-not-imp-rec}) that the Induction Axiom is consistent\index{consistent} with the negation of the Recursion Theorem.} \begin{example}\label{exam:ind-not-imp-rec} On $\B$, define a singulary operation $s$ by $s(0)=1$ and $s(1)=0$. Then $(\B,s,0)$ admits induction,\footnote{The structure $(\B,s,0)$ in Example \ref{exam:ind-not-imp-rec} also satisfies \axu, but not \axz. If we define $t:\B\to \B$ so that $t(n)=1$ for each $n$ in $\B$, then $(\B,t,0)$ satisfies the Induction Axiom and \axz, but not \axu. Later we shall have natural examples of structures satisfying \axz\ and \axu, but not admitting induction.} but there is \emph{no} function $g:\B\to\N$ such that $g(0)=0$ and $g(s(n))=\scr{(g(n))}$ for all $n$ in $\B$. \end{example} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove the Recursion Theorem by showing that, if $\family C$ is the set of all subsets $D$ of $A$ such that \begin{compactenum}[1)] \item $(0,b)\in D$, and \item if $(u,v)\in D$, then $(\scr u,f(v))\in D$, \end{compactenum} then $\bigcap\family C$ is the desired function $h$. \item Prove directly (without Theorem~\ref{thm:isom}) that \axz\ is a consequence of the Recursion Theorem. (For example, if in $\str A$ the successor-operation is surjective, show that there is no homomorphism from $\str A$ into $\N$.) \end{enumerate} \section{Arithmetic operations}\label{sect:arith-ops} By recursion, we can define addition, multiplication and exponentiation.\footnote{We can also define addition and multiplication using only the Induction Axiom, not the Recursion Theorem. The method is used by Landau~\cite{MR12:397m}. As a result, the operations can be defined on structures that do not satisfy all of the Peano Axioms. For example, let $n$ be a positive integer, and on $\Z$ let $\equiv$ be congruence \emph{modulo} $n$.% \index{congruence \emph{modulo} $n$}% \index{modulo@\emph{modulo}!congruence --- n@congruence --- $n$} if $x\equiv y$, then $x+1\equiv y+1$ (though by the standards of this chapter, we cannot quite prove this yet). Hence we can define a successor-operation $s$ on $\Z/\mathord\equiv$, namely $[x]\mapsto[x+1]$. The resulting structure $(\Z/\mathord\equiv,s,[0])$ satisfies the Induction Axiom; therefore it can be equipped with an addition and a multiplication that satisfy the theorems of this section. Thus we get arithmetic \textbf{\emph{modulo}} $n$. We can define exponentiation on $\Z/\mathord\equiv$ by $x^1=x$ and $x^{k+1}=x^k\cdot x$ if and only if $n$ is $1$, $2$, $6$, $42$, or $1806$. If we try the definition in case $n=3$, we get $2^1=2$, $2^2=2\cdot 2=1$, so $2^{s(2)}=2$, $2^{s(s(2))}=1$---but also $s(s(2))=1$, so $2^{s(s(2))}=2^1=2$.} First, we define the binary operation $+$ of \textbf{addition}\dindex{addition}\dindexsub{operation}{addition} on $\N$ by defining, for each $n$ in $\N$, the singulary operation $y\mapsto n+y$. This operation is given by the rules: \begin{compactenum}[1)] \item $n+0=n$; \item $n+\scr m=\scr{(n+m)}$. \end{compactenum} \begin{lemma}\label{lem:add} $\N$ satisfies \begin{compactenum}[1)] \item $\Forall x0+x=x$, \item $\Forall x\Forall y \scr y+x=\scr{(y+x)}$. \end{compactenum} \end{lemma} \begin{proof} By definition of addition, $0+0=0$. Suppose $0+n=n$. Then \begin{align*} 0+\scr n &=\scr{(0+n)}&&\text{[by definition of addition]}\\ &=\scr n.&&\text{[by inductive hypothesis]} \end{align*} This completes an induction showing $\models\Forall x0+x=x$. For the second claim, as the base step of an induction, we have \begin{align*} \scr m+0 &=\scr m&&\text{[by the first claim]}\\ &=\scr{(m+0)};&&\text{[again by the first claim]} \end{align*} so $\Forall y\scr y+0=\scr{(y+0)}$. Now, as an inductive hypothesis, suppose $\Forall y\scr y+n=\scr{(y+n)}$. Then, for all $m$ in $\N$, we have \begin{align*} \scr m+\scr n &=\scr{(\scr m+n)}&&\text{[by definition of addition]}\\ &=\scr{\scr{(m+n)}{}}&&\text{[by inductive hypothesis]}\\ &=\scr{(m+\scr n)}&&\text{[again by definition of addition]}. \end{align*} This completes an induction showing $\Forall x\Forall y\scr y+x=\scr{(y+x)}$. \end{proof} The second part of the proof showed $\N=\{x:\Forall y\scr y+x=\scr{(y+x)}\}$: We have proved the identity \begin{equation}\label{eqn:y'+x=(y+x)'} \scr y+x=\scr{(y+x)} \end{equation} in $\N$ by \textbf{induction on $x$}.\dindexsub{induction}{--- on x@--- on $x$} Induction on $y$ here does not work directly. Indeed, suppose $A=\{y\in\N:\Forall x\scr y+x=\scr{(y+x)}\}$. To prove that $0\in A$, we have to show that $\scr 0+n=\scr{(0+n)}$. From the first part of the theorem, we know that $\scr{(0+n)}=\scr n$; but we cannot yet say anything about $\scr 0+n$. We could prove $\Forall x\scr 0+x=\scr x$ by induction; but it would be more efficient just to start over and prove Identity~\eqref{eqn:y'+x=(y+x)'} by induction on $x$. To prove some identities below, one has to choose the right variable to work with. \begin{theorem}\label{thm:add} On $\N$, the following hold. \begin{compactenum} \item $\Forall x\scr x=x+1$. \item Addition is \textbf{commutative:}\dindex{commutative} \begin{equation*} \Forall x \forall y\qsep x+y=y+x. \end{equation*} \item Addition is \textbf{associative:}\dindex{associative} \begin{equation*} \Forall x \forall y\qsep \forall z\qsep (x+y)+z=x+(y+z). \end{equation*} \item Addition admits \textbf{cancellation:}\dindex{cancellation} \begin{equation*} \Forall x\Forall y\Forall z(x+z=y+z\lto x=y). \end{equation*} \end{compactenum} \end{theorem} We may henceforth write $n+1$ instead of $\scr n$. The \textbf{binomial coefficients}\dindexsub{binomial}{--- coefficient}\dindexsub{coefficient}{binomial ---} $\binom nm$ can be given recursively as follows. First we define $m\mapsto\binom0m$ by \begin{equation*} \binom0m= \begin{cases} 1,&\text{ if }m=0,\\ 0,&\text{ if }m\neq0. \end{cases} \end{equation*} Then, in terms of of $m\mapsto\binom nm$, we define $m\mapsto\binom{n+1}m$ recursively by \begin{equation*} \binom{n+1}m= \begin{cases} 1,&\text{ if }m=0,\\ \binom nk+\binom n{k+1},&\text{ if }m=k+1. \end{cases} \end{equation*} (See also Exercises \ref{exer:binom} and \ref{exer:bin-thm} in \S~\ref{order}.) The binary operation $\cdot$ of \textbf{multiplication}\dindex{multiplication}\dindexsub{operation}{multiplication} on $\N$ is given by: \begin{compactenum}[1)] \item $n\cdot 0=0$ \item $n\cdot(m+1)=n\cdot m+n$. \end{compactenum} Multiplication is also indicated by juxtaposition, so that $n\cdot m$ is $nm$. \begin{lemma}\label{lem:mult} $\N$ satisfies \begin{compactenum}[1)] \item $\Forall x 0x=0$, \item $\Forall x \forall y\qsep(y+1)x=yx+x$. \end{compactenum} \end{lemma} \begin{theorem}\label{thm:mult} On $\N$, the following hold. \begin{compactenum} \item $\Forall x 1x=x$. \item Multiplication is commutative ($\Forall x \forall y\qsep xy=yx$). \item Multiplication \textbf{distributes}\dindex{distributive} over addition: \begin{equation*} \Forall x \forall y\qsep \forall z\qsep (x+y)z=xz+yz. \end{equation*} \item Multiplication is associative ($\Forall x \forall y\qsep \forall z\qsep (xy)z=x(yz)$). \end{compactenum} \end{theorem} Finally, exponentiation: the binary operation $(x,y)\mapsto x^y$ on $\N$ is given by: \begin{compactenum}[1)] \item $n^0=1$; \item $n^{m+1}=n^m\cdot n$. \end{compactenum} \begin{theorem}\label{thm:exp} The following are identities\index{identity} in $\N$: \begin{compactenum}[1)] \item $x^{y+z}=x^yx^z$; \item $(x^y)^z=x^{yz}$; \item $(xy)^z=x^z y^z$. \end{compactenum} \end{theorem} \begin{proof} Exercise. \end{proof} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} %\item %Prove Lemma~\ref{lem:add}. \item Prove Theorem~\ref{thm:add}. In the latter two parts, does induction work on every variable? \item Prove that $\binom x1=x$ for all $x$ in $\N$. \item Prove Lemma~\ref{lem:mult}. In the second part, does induction work on either variable? \item Prove Theorem~\ref{thm:mult}. \item Prove Theorem~\ref{thm:exp}. \end{enumerate} \section{Rational numbers}\label{sect:ZandQ} \subsection*{The positive rational numbers} The integers can be constructed from the natural numbers, and the rational numbers can be constructed from the integers. However, the \emph{positive} rational numbers can also be constructed directly from the \emph{positive} natural numbers, and indeed we are taught some aspects of this construction from an early age. Let us denote the set of positive natural numbers, $\{1,2,3,\dots\}$, by \begin{equation*} \Np. \end{equation*} If $a$ and $b$ are natural numbers, then there is a \textbf{fraction}\index{fraction} denoted by \begin{equation*} \frac ab \end{equation*} or $a/b$. Then there are definitions for adding and multiplying fractions: \begin{align}\label{eqn:Q} \frac ab+\frac cd&=\frac{ad+cb}{bd},& \frac ab\cdot\frac cd&=\frac{ac}{bd}. \end{align} We are taught to \emph{reduce} fractions also: By~\eqref{eqn:Q} we compute $1/3+1/6=9/18$, which reduces to $1/2$. In particular, $9/18$ and $1/2$ are \emph{equal} fractions. Equality of fractions may be given by \begin{equation}\label{eqn:frac=} \frac ab=\frac cd\iff ad=cb. \end{equation} This equation is justified by: \begin{theorem}\label{thm:sim-eq} The relation $\sim$ on $\N\times\N$ is an equivalence-relation. \end{theorem} \begin{proof} Reflexivity and symmetry of $\sim$ follow immediately from the corresponding properties of equality; but transitivity needs more. Suppose $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$. Then $ad=cb$ and $cf=ed$, so \begin{equation*} (ad)f=(cb)f=c(bf)=c(fb)=(cf)b=(ed)b \end{equation*} by commutativity and associativity of multiplication. By these properties and also cancellation, we can go on to conclude \begin{equation*} af=eb, \end{equation*} hence $(a,b)\sim(e,f)$. \end{proof} The fraction $a/b$ is the equivalence-class $(a,b)/\mathord{\sim}$, where \begin{equation}\label{eqn:sim} (a,b)\sim(x,y)\iff ay=bx. \end{equation} Let us denote $(\Np\times\Np)/\mathord{\sim}$ by \begin{equation*} \Qp. \end{equation*} This is the set of \textbf{positive rational numbers.}\index{positive rational numbers} \subsubsection*{Structure} We are free to define operations $\oplus$ and $\otimes$ on $\N\times\N$ by \begin{align*} (a,b)\oplus(c,d)&=(ad+cb,bd),& (a,b)\otimes(c,d)&=(ac,bd). \end{align*} What makes these useful is the following: \begin{theorem}\label{thm:+.-on-Nmod} If $a/b=a'/b'$ and $c/d=c'/d'$, then \begin{gather}\notag (a,b)\oplus(c,d)\sim(a',b')\oplus(c',d'),\\\notag (a,b)\otimes(c,d)\sim(a',b')\otimes(c',d'). \end{gather} \end{theorem} \begin{corollary}\label{cor:+.-on-Nmod} On $\Qp$, the equations~\eqref{eqn:Q} define two binary operations. \end{corollary} \begin{theorem}\label{thm:Q^+} On $\Qp$, \begin{compactenum}[1)] \item addition and multiplication are commutative and associative, \item multiplication distributes over addition, \item $1$ is a \textbf{multiplicative identity:}% \dindexsub{identity}{multiplicative ---}% \dindexsub{multiplicative}{--- identity} \begin{equation*} \Forall x1\cdot x=x. \end{equation*} \end{compactenum} \end{theorem} The whole point of defining $\Qp$ is the following: \begin{theorem}\label{thm:sol-in-Q+} There is a well-defined operation $x\mapsto x\inv$ on $\Qp$ given by \begin{equation*} \left(\frac ab\right)\inv=\frac ba. \end{equation*} This operation is \textbf{multiplicative inversion:}% \dindexsub{inversion}{multiplicative}% \dindexsub{multiplicative}{--- inversion} \begin{equation*} \Forall xx\cdot x\inv=1. \end{equation*} \end{theorem} Therefore, if $r$ and $s$ are in $\Qp$, then the equation $r=s\cdot x$ has the unique solution $s\inv r$, which is written also as a fraction, \begin{equation*} \frac rs. \end{equation*} If $a,b,c,d\in\N$, then \begin{equation*} \frac{a/b}{c/d}=\frac{ad}{bc}, \end{equation*} and in particular \begin{equation}\label{eqn:a/1/c/1} \frac{a/1}{c/1}=\frac ac. \end{equation} \subsubsection*{Numbers and fractions} By our construction, a positive natural number is not literally a positive rational number; a positive rational number is a class of ordered pairs of positive natural numbers. One way to understand this is shown in Figure~\ref{fig:grid}, \begin{figure}[t] \begin{center} \psset{unit=3mm} \begin{pspicture}(12,12) \psgrid[subgriddiv=1,gridlabels=0,griddots=4](12,12)(12,12)(0,0) \uput[r](12,12){$O$} \psline(-1,3.333)(12,12) \uput[l](-1,3.333){$\displaystyle\frac32$} \psdots(9,10)(6,8)(3,6)(0,4) \psset{framesep=0pt} \uput[dr](9,10){\psframebox*{$(3,2)$}} \uput[dr](6,8){\psframebox*{$(6,4)$}} \uput[dr](3,6){\psframebox*{$(9,6)$}} \uput[dr](0,4){\psframebox*{$(12,8)$}} \psdots[dotstyle=o](12,12) \end{pspicture} \end{center} \caption{Fractions as straight lines}\label{fig:grid} \end{figure} where ordered pairs of natural numbers are depicted as points in a grid; then a fraction is the class of ordered pairs lying on a particular straight line through the point $O$. A fraction may not literally be a positive natural number; but there are fractions that \emph{behave} like natural numbers: \begin{theorem}\label{thm:N-in-Q+} The function $x\mapsto x/1$ is an embedding\index{embedding} of $(\Np,1,+,\cdot)$ in $(\Qp,1/1,+,\cdot)$; that is, it is injective, it takes $1$ to $1/1$, and \begin{align*} \frac{x+y}1&=\frac x1+\frac y1,& \frac{x\cdot y}1&=\frac x1\cdot\frac y1. \end{align*} \end{theorem} \begin{proof} Immediate from the definitions. \end{proof} We may therefore forget about the distinction between natural numbers and positive rational numbers: we may \emph{identify} a natural number $n$ with its image $n/1$ in $\Qp$. By~\eqref{eqn:a/1/c/1}, there will be no ambiguity in writing fractions: a fraction of natural numbers as such will be the same as their fraction as positive rational numbers. Using the idea in Figure~\ref{fig:grid}, we can arrange the positive rational numbers along a semicircle, according to their ordering, as in Figure~\ref{fig:Q+}~\eqref{item:circ}. \begin{figure}[t] \mbox{}\hfill \begin{inparaenum}[(a)] \item\label{item:circ} \begin{pspicture}[](-4.5,-4.5)(0,0) \psset{unit=1.5cm} \psgrid[subgriddiv=1,gridlabels=0,griddots=12](0,0)(-3,-3) \psarc(0,-1){1}{90}{270} \uput[l](-1,-1){\psframebox*{$1$}} \psline(-3,-2)(0,0)(-2,-3) \psdots(-0.923,-0.615)(-0.923,-1.385)(-1,-1) \uput[l](-0.923,-0.615){$3/2$} \uput[l](-0.923,-1.385){$2/3$} \psdots[dotstyle=o](0,0) \end{pspicture} \hfill \item\label{item:line} \begin{pspicture}[](-4.5,-4.5)(0,0) \psset{unit=1.5cm} \psgrid[subgriddiv=1,gridlabels=0,griddots=12](0,0)(-3,-3) \psline(-3,-2)(0,0)(-2,-3) \psline(-3,-1)(0,-1) \psdots(-1.5,-1)(-0.667,-1)(-1,-1) \uput[ul](-1.5,-1){$3/2$} \uput[dl](-1,-1){$1$} \uput[dr](-0.667,-1){$2/3$} \psdots[dotstyle=o](0,0) \end{pspicture} \end{inparaenum} \hfill\mbox{} \caption{Positive rationals along a semicircle and a straight line}\label{fig:Q+} \end{figure} It is more usual to arrange the positive rational numbers along a straight line, as in Figure~\ref{fig:Q+}~\eqref{item:line}; the point of using a semicircle is that here, if $k0$, then $z=w+y$, so $x\cdot z=x\cdot w+x\cdot y$, and therefore \begin{equation*} x\cdot(-y+z)=-(x\cdot y)+x\cdot z=x\cdot-y+x\cdot z.\qedhere \end{equation*} \end{proof} \subsection*{The rational numbers} As we obtained $\Z$ from $\N$, so we can obtain $\Z$ from $\Np$. The difference is that the embedding of $\Np$ in $\Z$ is $x\mapsto(x+1)-1$, and $0$ in $\Z$ is $1-1$. We can now obtain $(\Q,0,-,+,\cdot)$ from $(\Qp,+,\cdot)$ just as we obtain $(\Z,0,-,+,\cdot)$ from $(\Np,+,\cdot)$. \begin{theorem}\label{thm:Q} Addition and multiplication are commutative and associative on $\Q$, and multiplication distributes over addition. Addition has the identity $0$, and multiplication has the identity $1$. The operation $x\mapsto-x$ is additive inversion, and there is an operation $x\mapsto x\inv$ of multiplicative inversion on $\Q\setminus\{0\}$. \end{theorem} Because of this theorem, $\Q$ is called a \textbf{field.} \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item Prove Theorem~\ref{thm:+.-on-Nmod} and Corollary~\ref{cor:+.-on-Nmod}. \item Prove Theorem~\ref{thm:Q^+}. \item Prove Theorem~\ref{thm:sol-in-Q+}. \item Prove Theorem~\ref{thm:approx-eq}. \item Prove Theorem~\ref{thm:Z-+}. \item Prove Theorem~\ref{thm:Z-ocm}. \item Prove Theorem~\ref{thm:Z-}. \item Prove Theorem~\ref{thm:N-in-Z}. \item Prove Theorem~\ref{thm:Z+.}. \item Prove Theorem~\ref{thm:Q}. \end{enumerate} \section{More recursion}\label{sect:recursion-gen} Informally, we define $n!$\glossary{$n!$}, that is, $n$\textbf{-factorial,}% \dindexsub{factor}{---ial}% \dindex{nfactorial@$n$-factorial} by \begin{equation*} n!=1\cdot 2\cdot 3\cdots(n-1)\cdot n. \end{equation*} More precisely, we have the recursive definition \begin{align}\label{eqn:rec-gen} 0!&=1,&(n+1)!&=n!\cdot(n+1). \end{align} However, for this to be a valid definition by the Recursion Theorem as it is, we would have to express $n!\cdot(n+1)$ as a function of $n!$. Alternatively,~\eqref{eqn:rec-gen} is a valid recursive definition by the following. \begin{theorem}[Recursion with Parameter]\label{thm:rec-param}\dindexsub{recursion}{R--- Theorem with Parameter}\dindexsub{theorem}{Recursion Th--- with Parameter}\dindexsub{parameter}{Recursion Theorem with P---} Suppose $B$ is a set with an element $c$, and $F:\N\times B\to B$. Then there is a \emph{unique} function $G$ from $\N$ to $B$ such that $G(0)=c$ and \begin{equation}\label{eqn:str-rec} G(n+1)=F(n,G(n)) \end{equation} for all $n$ in $\N$. \end{theorem} \begin{proof} Let $f$ be the function \begin{equation*} (x,b)\longmapsto(x+1,F(x,b)) \end{equation*} from $\N\times B$ to $\N\times B$. By recursion, there is a unique function $g$ from $\N$ to $\N\times B$ such that $g(0)=(0,c)$ and $$g(n+1)=f(g(n))$$ for all $n$ in $\N$. Now let $G$ be $\pi\circ g$, where $\pi$ is the function \begin{equation*} (x,b)\longmapsto b \end{equation*} from $\N\times B$ to $B$. Then for each $n$ in $\N$ we have $g(n)=(m,G(n))$ for some $m$ in $\N$. We can prove by induction that $m=n$. Indeed, this is the case when $n=0$, since $g(0)=(0,c)$. Suppose $g(n)=(n,G(n))$ for some $n$ in $\N$. Then \begin{equation}\label{eqn:strong-rec} g(n+1)=f(n,G(n))=(n+1,F(n,G(n))). \end{equation} In particular, the first entry in the value of $g(n+1)$ is $n+1$. This completes our induction. We now know that $g(n)=(n,G(n))$ for all $n$ in $\N$. Hence in particular $g(n+1)=(n+1,G(n+1))$. But we also have (\ref{eqn:strong-rec}). Therefore we have (\ref{eqn:str-rec}), as desired. Finally, each of $g$ and $G$ determines the other. Since $g$ is unique, so is $G$. \end{proof} \begin{example} We can define a function $f$ on $\N$ by requiring $f(0)=0$ and $f(x+1)=x$. This is a valid recursive definition, by Theorem \ref{thm:rec-param}. Note that $f$ picks out the immediate predecessor of a natural number, when this exists.\footnote{Since $f$ is unique, we now have a proof that \axu\ follows from the Recursion Theorem.} \end{example} For any function $f$ from $\Np$ to $M$, where $M$ is a set equipped with addition and multiplication, we can now define the sum $\sum_{k=1}^nf(k)$ and the product $\prod_{k=1}^nf(k)$ recursively as follows: \begin{align*} \sum_{k=1}^0f(k)&=0,&\sum_{k=1}^{n+1}f(k)&=\sum_{k=1}^nf(k)+f(n+1),\\ \prod_{k=1}^0f(k)&=1,&\prod_{k=1}^{n+1}f(k) &=\prod_{k=1}^nf(k)\cdot f(n+1). \end{align*} See Exercise~\ref{exercise:SP} below. \subsection*{Exercises} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi.} \item\label{exercise:SP} Show clearly that the definitions of $\sum_{k=1}^nf(k)$ and $\prod_{k=1}^nf(k)$ are justified by Theorem~\ref{thm:rec-param}. \item Prove the following for all $n$ in $\N$: \begin{enumerate} \item $\sum_{k=1}^nk=n(n+1)/2$; \item $\sum_{k=1}^nk^2=n(n+1)(2n+1)/6$; \item $\sum_{k=1}^nb^{k-1}=(b^n-1)/(b-1)$; \item $\sum_{k=1}^n(2k-1)=n^2$; \item $\prod_{k=1}^n(k/(k+1))=1/(n+1)$. \end{enumerate} \end{enumerate} \section{Ordering of the natural numbers}\label{order} We can define the binary relation $\leq$ on $\N$ as the set \begin{equation*} \{(x,y)\in\N\times \N:\Exists zx+z=y\}. \end{equation*} The associated strict relation $<$ is then $\{(x,y)\in\N\times\N:x\leq y\land x\neq y\}$. Now we have to show that $\leq$ is the linear ordering that we expect: \begin{lemma}\label{lem:cancel} $\N\models\Forall x\Forall y(x+1\leq y+1\lto x\leq y)$. \end{lemma} \begin{proof} Suppose $a+1\leq b+1$. Then $a+1+c=b+1$ for some $c$ in $\N$, by definition of $\leq$. This means $a+c+1=b+1$, by Lemma~\ref{lem:add}, so $a+c=b$, by \axu, and therefore $a\leq b$, again by the definition of $\leq$. \end{proof} \begin{lemma}\label{lem:x-leq-0} $\N$ satisfies: \begin{compactenum}[1)] \item $\Forall x(x\leq0\lto x=0)$; \item $\Forall x\Forall y(x+y\leq x\lto y=0)$. \end{compactenum} \end{lemma} \begin{proof} Suppose $a\leq0$. Then $a+b=0$ for some $b$ in $\N$. Either $a=0$, or $a=c+1$ for some $c$ in $\N$, by Lemma~\ref{lem:zero-succ}. In the latter case, $c+b+1=0$, which is absurd by \axz. Hence $a=0$, and the first claim is proved. Now suppose $a+b\leq a$. Then $a+b+c=a=a+0$ for some $c$, so $b+c=0$ by cancellation (Theorem~\ref{thm:add}), which means $b\leq0$. Hence $b=0$ by the first claim. The second claim is now proved. \end{proof} \begin{lemma}\label{lem:<1} $\N$ satisfies: \begin{compactenum}[1)] \item $\Forall x\Forall y(x