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\newtheorem{exercise}{Al\i\c st\i rma}

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 \begin{document}
 \title{\"Onermeler mant\i\u g\i ndaki\\
  bi\c cimsel kan\i tlar}
 \author{David Pierce}
 \date{\today, \printtime}
 \maketitle

%Bu belge \c su anda son hali de\u gildir.  Sonra bak\i n.

%\section*{\"Ons\"oz}

Bu yaz\i n\i n ana
kaynaklar\i, Burris'in \cite{Burris} ve Nesin'in \cite{Nesin-OM}
kitaplar\i\ ve \foreignlanguage{english}{\emph{Foundations of Mathematical Practice}} (Eyl\"ul 2010)
adl\i\ notlar\i m.
Baz{\i} terimler, \cite{Demirtas,MTS}
 kaynaklar{\i}ndan al{\i}nm\i\c st{\i}r.

 \tableofcontents

\section{\"Onermeler}

 \textbf{\"Onerme,} belli bir \emph{durumda}
 \emph{do\u gru} veya \emph{yanl{\i}\c s} denebilen \emph{c\"umledir.} 
Matematikte, durum \c co\u gunlukla bir \emph{yap{\i}d{\i}r.}
\"Orne\u gin, \emph{her say{\i}n{\i}n tersi var} c\"umlesi, bir \"onermedir, ve bu \"onerme,
\begin{compactitem}
\item
$(\N,+)$ yap{\i}s{\i}nda yanl{\i}\c s,
\item
$(\Z,+)$ yap{\i}s{\i}nda do\u gru,
\item
$(\Z,\cdot)$ yap{\i}s{\i}nda yanl{\i}\c s,
\item
$(\Qp,\cdot)$ yap{\i}s{\i}nda do\u grudur.  (Burada $\Qp=\{x\colon x\in\Q\land x>0\}$.)
\end{compactitem}

Belli bir durumda, bir \"onermenin \textbf{do\u gruluk de\u geri} vard{\i}r:
\begin{compactitem}
\item 
\"onerme do\u gru ise, de\u geri $1$;
\item
\"onerme yanl{\i}\c s ise, de\u geri $0$'d{\i}r.
\end{compactitem}
$\{0,1\}$ k\"umesi i\c cin,
\begin{equation*}
  \B
\end{equation*}
harf{}ini kullanal{\i}m.
\"Onermeler i\c cin, $P$, $Q$, ve $R$ gibi Latin harflerini kullanal{\i}m.
Sonsuz tane \"onermemiz varsa, $P_0$, $P_1$, $P_2$ ve benzerlerini kullanabiliriz.
Bir $d$ durumunda, $P$ \"onermesinin $\B$ k\"umesindeki de\u geri olarak
\begin{equation*}
  d(P)
\end{equation*}
kullan{\i}labilir.  O zaman $d$, \"onermeler k\"umesinden $\B$ k\"umesine bir
fonksiyondur (d\"on\"u\c s\"umd\"ur):
\begin{equation*}
  d\colon\{\text{\"onermeler}\}\to\B.
\end{equation*}

\section{Do\u gruluk tablolar\i}

Verilmi\c s \"onermelerden, \textbf{ba\u gla\c clarla, bile\c ske \"onermeler}
yap{\i}labilir, ve onlar{\i}n de\u gerleri, verilmi\c s \"onermelerin de\u gerlerinden
bulunabilir.  \"Orne\u gin:
\begin{center}
  $P$ ve $Q$;\\
$P$ veya $Q$;\\
$P$ ise $Q$;\\
$P$ ancak ve ancak $Q$;\\
$P$ de\u gil.
\end{center}
(Dilbilgisinde, \emph{ise} ve \emph{de\u gil,} ba\u gla\c c de\u gildir; ama matematikte, \"oyle say{\i}labilir.)
Bu \"orneklerde, s\"ozc\"uklerin yerinde, k{\i}saltma olarak, simgeler
kullan{\i}labilir:\footnote{$P\land Q$ yerine, $P\mathrel{\&}Q$; $P\lto Q$ yerine, $P\to Q$; $P\liff P$ yerine, $P\leftrightarrow Q$, yaz\i labilir.}
\begin{gather*}
  P\land Q\\
P\lor Q\\
P\lto Q\\
P\liff Q\\
\lnot P
\end{gather*}
Yukar\i daki simgelere, \textbf{ba\u glay\i c\i}  diyelim.
Bile\c ske \"onermelerin farkl\i\ durumlardaki do\u gruluk de\u gerleri, \textbf{do\u gruluk tablolar\i nda}\footnote{Veya do\u gruluk \c cizelgelerinde.}
g\"osterilebilir:
\begin{align*}
&  \begin{array}{cc|cccc}
    P&Q&P\land Q&P\lor Q&P\lto Q&P\liff Q\\\hline
0&0&0&0&1&1\\
1&0&0&1&0&0\\
0&1&0&1&1&0\\
1&1&1&1&1&1
  \end{array}&
&
  \begin{array}{c|c}
    P&\lnot P\\\hline
0&1\\
1&0
  \end{array}
\end{align*}
Mesela, ikinci sat\i rdan, $d(P)=1$ ve $d(Q)=0$ ise, o zaman $d(P\lto Q)=0$.

Do\u gruluk tablolar\i\ \c s\"oyle yaz{\i}labilir:
\begin{align*}
  &
  \begin{array}{ccc}
P&\land&Q\\\hline
0&0&0\\
1&0&0\\
0&0&1\\
1&1&1
  \end{array}&
&
  \begin{array}{ccc}
P&\lor&Q\\\hline
0&0&0\\
1&1&0\\
0&1&1\\
1&1&1
  \end{array}&
&
  \begin{array}{ccc}
P&\lto&Q\\\hline
0&1&0\\
1&0&0\\
0&1&1\\
1&1&1
  \end{array}&
&
  \begin{array}{ccc}
P&\liff&Q\\\hline
0&1&0\\
1&0&0\\
0&0&1\\
1&1&1
  \end{array}&
&
  \begin{array}{cc}
\lnot&P\\\hline
0&1\\
1&0
  \end{array}
\end{align*}
Burada, bir \"onermenin de\u geri, \"onermenin ba\u glay\i c\i s\i n\i n alt\i na yaz\i l\i r.

Asl{\i}nda, $P\lto Q$ gibi bir ifade ger\c cek bir \"onerme de\u gildir; bir \textbf{\"onerme
  form\"ul\"ud\"ur;} ve $P$ ve $Q$ harfleri, \textbf{\"onerme de\u gi\c skenleridir.}
De\u gi\c skenlerin yerine ger\c cek \"onermeleri 
koyarsak, \"onerme form\"ul\"u, bir \"onerme olur.

Bir \"onerme form\"ul\"unde, birden fazla ba\u glay\i c\i\
kullan{\i}labilir, mesela
\begin{equation*}
  P\lor\lnot P
\end{equation*}
form\"ul\"unde oldu\u gu gibi.
Bu form\"ul\"un, \"u\c c veya d\"ort tane \textbf{altform\"ul\"u} var; bunlar
\begin{align*}
  &P\lor\lnot P,&&P,&&\lnot P,&&P
\end{align*}
form\"ulleridir.
Biz, $P$ altform\"ul\"un\"u tekrarlad{\i}k, \c c\"unk\"u form\"ul\"u bir \textbf{a\u ga\c c} \c seklinde g\"orebiliriz,
ve bu a\u gac{\i}n d\"ort tane \textbf{d\"u\u g\"um\"u} olur:
\begin{equation*}
  \xymatrix%@!0%
{
 & *+[F]{P\lor \lnot P} \ar@{-}[dl] \ar@{-}[dr]  &&\\
*+[F]{P}& & *+[F]{\lnot P} \ar@{-}[dr] &\\
 & &                     & *+[F]{P}
}
\end{equation*}
A\u ga\c cta, altform\"ul\"un yerine, form\"ul\"un \textbf{ana ba\u glay\i c\i s\i n\i}
koyabiliriz:
\begin{equation*}
  \xymatrix%@!0
{
 & *+[F]{\lor} \ar@{-}[dl] \ar@{-}[dr]  &&\\
*+[F]{P}& & *+[F]{\lnot} \ar@{-}[dr] &\\
 & &                     & *+[F]{P}
}
\end{equation*}
O zaman $P\lor\lnot P$ form\"ul\"un do\u gruluk tablosu iki \c sekilde yaz{\i}labilir:
\begin{align*}
&
  \begin{array}{c|c|c}
    P&\lnot P&P\lor\lnot P\\\hline
0&1&1\\
1&0&1
  \end{array}
&&
  \begin{array}{cccc}
    P&\lor&\lnot&P\\\hline
0&1&1&0\\
1&1&0&1
  \end{array}
\end{align*}
\.Ikinci \c sekilde, her altform\"ul\"un de\u geri, altform\"ul\"un ana ba\u glay\i c\i s\i n\i n
alt{\i}na yaz{\i}l{\i}r.

Bir \"ornek  daha:  $P\lto\lnot Q\lor R$ (yani, $P\lto((\lnot Q)\lor
R)$) \"onerme form\"ul\"un\"un a\u gac{\i} a\c sa\u g\i dad\i r:
\begin{equation*}
  \xymatrix%@!0
{
  & *+[F]{P\lto \lnot Q\lor R}\ar@{-}[dl] \ar@{-}[drrr]&&&&\\
*+[F]{P}&&&&*+[F]{\lnot Q\lor R}\ar@{-}[dll] \ar@{-}[dr] &\\
&& *+[F]{\lnot Q}\ar@{-}[dr] &&&*+[F]{R}\\
&&&*+[F]{Q}&&
}
\end{equation*}
Ana ba\u glay\i c\i lar\i\ kullanarak,
\begin{equation*}
\xymatrix%@!0
{
  & *+[F]{\lto}\ar@{-}[dl] \ar@{-}[drrr]&&&&\\
*+[F]{P}&&&&*+[F]{\lor}\ar@{-}[dll] \ar@{-}[dr] &\\
&& *+[F]{\lnot}\ar@{-}[dr] &&&*+[F]{R}\\
&&&*+[F]{Q}&&
}
\end{equation*}
\c seklinde yazabiliriz.
Do\u gruluk tablosu \c s\"oyle yaz{\i}labilir:
\begin{align*}
&\begin{array}{c|c|c|c|c|c}
 P&\lto&\lnot&Q&\lor&R\\ \hline
 0&&&0&&0\\
 1&&&0&&0\\
 0&&&1&&0\\
 1&&&1&&0\\
 0&&&0&&1\\
 1&&&0&&1\\
 0&&&1&&1\\
 1&&&1&&1
\end{array},
&&
\begin{array}{c|c|c|c|c|c}
 P&\lto&\lnot&Q&\lor&R\\ \hline
 0&&1&0&&0\\
 1&&1&0&&0\\
 0&&0&1&&0\\
 1&&0&1&&0\\
 0&&1&0&&1\\
 1&&1&0&&1\\
 0&&0&1&&1\\
 1&&0&1&&1
\end{array},
&&
\begin{array}{c|c|c|c|c|c}
 P&\lto&\lnot&Q&\lor&R\\ \hline
 0&&1&0&1&0\\
 1&&1&0&1&0\\
 0&&0&1&0&0\\
 1&&0&1&0&0\\
 0&&1&0&1&1\\
 1&&1&0&1&1\\
 0&&0&1&1&1\\
 1&&0&1&1&1
\end{array},
\end{align*}
ve sonunda,
\begin{equation*}
\begin{array}{c|c|c|c|c|c}
 P&\lto&\lnot&Q&\lor&R\\ \hline
 0&1&1&0&1&0\\
 1&1&1&0&1&0\\
 0&1&0&1&0&0\\
 1&0&0&1&0&0\\
 0&1&1&0&1&1\\
 1&1&1&0&1&1\\
 0&1&0&1&1&1\\
 1&1&0&1&1&1
\end{array}.
\end{equation*}
O zaman form\"ul\"un \textbf{basit} do\u gruluk tablosu \c s\"oyledir:
\begin{equation*}
\begin{array}{ccc|c}
P&Q&R&P\lto\lnot Q\lor R\\\hline
0&0&0&1\\
1&0&0&1\\
0&1&0&1\\
1&1&0&0\\
0&0&1&1\\
1&0&1&1\\
0&1&1&1\\
1&1&1&1
\end{array}
\end{equation*}

Genellikle do\u gruluk de\u gerleri \c su s{\i}rada hesaplan{\i}r:
\begin{compactenum}
\item 
$\lnot$
\item 
$\land$ ve $\lor$
\item 
$\lto$ ve $\liff$
\item 
bir ba\u glay\i c\i\ iki kere kullan\i lm\i\c ssa,
sa\u gdaki.
\end{compactenum}
\"Orne\u gin:
  \begin{compactenum}
    \item
$P\lto Q\lor R$ demek $P\lto(Q\lor R)$;
\item
$\lnot P\land Q$ demek $(\lnot P)\land Q$;
\item
$P\land Q\lor R$ belirsiz;
\item
$P\land Q\land R$ demek $P\land(Q\land R)$;
\item
$P\land Q\land R\lor P$ belirsiz;
\item
$P\lto Q\lto R$ demek $P\lto(Q\lto R)$;
\item
$P\lto Q\land R\lto S$ demek $P\lto ((Q\land R)\lto S)$. 
  \end{compactenum}
$\land$, $\lor$, $\lto$, $\liff$ ba\u glay\i c\i lar\i na \textbf{ikili} denir;
  $\lnot$ ba\u glay\i c\i s\i na, \textbf{birli} denir.  Ayr{\i}ca, $0$ ve $1$, \textbf{s{\i}f{\i}rl{\i}
    ba\u glay\i c\i lar} olarak d\"u\c s\"un\"ulebilir.

\begin{exercise}
A\c sa\u g{\i}daki form\"ulleri hesaplay{\i}n.
\begin{compactenum}
\item
$1\lto 1\lto 1$,
\item
$1\lto 0\lto 1$,
\item
$(0\lto 1)\liff 1$,
\item
$(0\liff 1)\liff(0\liff 1)$,
\item
$\lnot\lnot\lnot 0$,
\item
$(1\lor 0)\land 0$,
\item
$1\lor (0\land 0)$.
\end{compactenum}
\end{exercise}

\begin{exercise}
A\c sa\u g{\i}daki form\"ullerin do\u gruluk tablolar\i n\i\ yap{\i}n:
\begin{compactenum}
\renewcommand{\labelenumii}{(\theenumii)}
\item
$P\lto Q\lto P$;
\item
$P\land Q\land R$;
\item
$\lnot(P\liff\lnot(Q\liff R))$;
\item
$(P\lto Q\lor R)\lto\lnot P\lor Q$;
\item
 $(P\lto Q\lor\lnot R)\land(Q\lto P\land R)\lto P\lto R$;
\item
$\lnot(\lnot R\lto P\lto\lnot(R\lto Q))$.
\end{compactenum}
\end{exercise}

\section{E\c sde\u gerlik}

\.Iki \"onermenin do\u gruluk de\u geri her durumda ayn\i ysa, o
\"onermeler, mant\i ksal olarak \textbf{e\c sde\u ger} veya
\textbf{denktir.}  \.Iki \"onerme \emph{form\"ul\"un\"un basit} do\u gruluk
tablolar\i\ ayn\i ysa, o form\"ulleri de birbirine \textbf{e\c sde\u
  ger} veya \textbf{denktir.}  $F$ ve $G$ \"onerme form\"ulleri e\c
sde\u ger ise,   
\begin{equation*}
F\sim G
\end{equation*}
ifadesini 
yazal\i m.
\"Orne\u gin,
\begin{equation*}
P\lto Q\sim \lnot P\lor Q
\end{equation*}
denkli\u gini a\c sa\u g\i daki tablolardan g\"orebiliriz:
\begin{align*}
&
  \begin{array}{ccc}
P&\lto&Q\\\hline
0&1&0\\
1&0&0\\
0&1&1\\
1&1&1
  \end{array}&
&
  \begin{array}{cccc}
\lnot&P&\lor&Q\\\hline
1&0&1&0\\
0&1&0&0\\
1&0&1&1\\
0&1&1&1
  \end{array}
\end{align*}
Buradan basit do\u gruluk tablolar\i n\i n ayn\i\ oldu\u gunu g\"or\"ur\"uz:
\begin{align*}
&
\begin{array}{cc|c}
P&Q&P\lto Q\\\hline
0&0&1\\
1&0&0\\
0&1&1\\
1&1&1
\end{array}
&&
\begin{array}{cc|c}
P&Q&\lnot P\lor Q\\\hline
0&0&1\\
1&0&0\\
0&1&1\\
1&1&1
\end{array}
\end{align*}

\begin{theorem}\label{thm:denk}
A\c sa\u g\i daki e\c sde\u gerliklerimiz vard\i r.
  \begin{compactenum}
\item
(Her \"onerme, sadece $\lnot$ ve $\land$ ile yaz\i labilir:)
  \begin{align*}
  P\lor Q&\sim\lnot(\lnot P\land\lnot Q),\\
    {\sv P}\lto {\sv Q}&\sim \lnot {\sv P}\lor {\sv Q},\\
{\sv P}\liff {\sv Q}&\sim({\sv P}\lto {\sv Q})\land ({\sv Q}\lto {\sv P}).
  \end{align*}
  \item
  (Her \"onerme, sadece $\lnot$ ve $\lto$ ile yaz\i labilir:)
\begin{equation*}
  P\land Q\sim\lnot(P\lto\lnot Q).
  \end{equation*}
    \item
    (\c Cifte de\u gilleme kald\i r\i labilir:)
\begin{equation*}
  \lnot\lnot {\sv P}\sim {\sv P}.
\end{equation*}
\item
(De Morgan\footnote{Augustus De Morgan, 1806--71, B\"uy\"uk Britanyal\i\ matematik\c ci ve mant\i k\c c\i\ \cite{Struik,Struik-TR}.} kurallar\i:)
\begin{align*}
  \lnot ({\sv P}\lor {\sv Q})&  \sim \lnot {\sv P}\land \lnot {\sv Q},\\
  \lnot ({\sv P}\land {\sv Q})&  \sim \lnot {\sv P}\lor \lnot {\sv Q}.
\end{align*}
\item
($\land$ ve $\lor$ ba\u glay\i c\i lar\i n\i n de\u gi\c sme ve birle\c sme \"ozellikleri:)
\begin{align*}
  {\sv P}&\land {\sv Q}  \sim {\sv Q}\land {\sv P},&
  ({\sv P}&\land {\sv Q})\land {\sv R}  \sim {\sv P}\land ({\sv Q} \land {\sv R}),\\
  {\sv P}&\lor {\sv Q}  \sim {\sv Q}\lor {\sv P},&
  ({\sv P}& \lor {\sv Q})\lor {\sv R}  \sim {\sv P}\lor ({\sv Q}\lor {\sv R}).
\end{align*}
\item
($\land$ ve $\lor$ ba\u glay\i c\i lar\i\ birbirine \"uzerine da\u g\i l\i r:)
\begin{align*}
 {\sv P}&\land({\sv Q}\lor {\sv R})\sim ({\sv P}\land {\sv Q})\lor({\sv P}\land {\sv R}),\\
 {\sv P}&\lor({\sv Q}\land {\sv R})\sim ({\sv P}\lor {\sv Q})\land({\sv P}\lor {\sv R}).
\end{align*}
\item
(Fazlal\i klar:)
\begin{align*}
  {\sv P}\land {\sv P} & \sim {\sv P}, & {\sv P}\land\lnot {\sv P} & \sim 0, & {\sv P}\land 1 & \sim {\sv P}, &
  {\sv P}\land 0 & \sim 0,\\ 
  {\sv P}\lor  {\sv P} & \sim {\sv P}, & {\sv P}\lor \lnot {\sv P} & \sim 1, & {\sv P}\lor  0 & \sim {\sv P}, &
  {\sv P}\lor  1 & \sim 1.
\end{align*}
\item
(Yeni de\u gi\c sken:)
\begin{align*}
  {\sv P}&\sim ({\sv P}\land {\sv Q})\lor ({\sv P}\land \lnot {\sv Q}),\\
  {\sv P}&\sim ({\sv P}\lor {\sv Q})\land ({\sv P}\lor \lnot {\sv Q}).
\end{align*}
\item
(Yutma:)
\begin{align*}
{\sv P}\land({\sv P}\lor {\sv Q})& \sim {\sv P},\\
{\sv P}\lor({\sv P}\land {\sv Q})& \sim {\sv P}.
\end{align*}
  \end{compactenum}
\end{theorem}

\begin{proof}
Al\i\c st\i rma.
\end{proof}

\section{Gerektirme}

Bir durum, bir \"onermeler k\"umesinin \textbf{modelidir,} e\u ger o
durumda, k\"umenin her \"onermesi do\u gru ise. 
E\u ger bir \"onerme, bir \"onermeler k\"umesinin her modelinde do\u
gru ise, o k\"ume, \"onermeyi \textbf{gerektirir.} 

$\Gamma$ (yani, \emph{Gamma}), bir \"onerme form\"ul\"u k\"umesi
olsun, ve $F$, bir \"onerme form\"ul\"u olsun.  E\u ger
$\Gamma\cup\{F\}$ k\"umesindeki b\"ut\"un form\"ullerin do\u gruluk
tablosunun her sat\i r\i nda,
\begin{compactenum}
\item 
ya $\Gamma$ k\"umesindeki bir form\"ul yanl\i\c s ise,
\item
ya da $F$ form\"ul\"u do\u gru ise,
\end{compactenum}
o zaman $\Gamma$, $F$ form\"ul\"un\"u \textbf{gerektirir} deriz,
ve
\begin{equation*}
\Gamma\models F
\end{equation*}
ifadesini
yazar\i z; $\models$ simgesine, \textbf{turnike} denebilir.  Bo\c
s k\"ume, $F$ form\"ul\"un\"u gerektirirse, 
\begin{equation*}
\models F
\end{equation*}
yaz\i labilir, ve $F$ form\"ul\"une \textbf{do\u grusal ge\c cerli
  form\"ul,} veya \textbf{mant\i ksal do\u gru form\"ul,} veya
\textbf{totoloji} denebilir. 

\begin{theorem}
$F\sim G$ ancak ve ancak $F\liff G$ bir totolojidir.
\end{theorem}

\begin{proof}
Al\i\c st\i rma.
\end{proof}

$\Gamma$ k\"umesinin $F$ form\"ul\"un\"u gerektirdi\u gini nas\i l
g\"osterebiliriz?  \.Iki y\"ontemimiz var: 
\begin{compactenum}
\item
do\u gruluk tablolar\i,
\item
\emph{bi\c cimsel kan\i t.}
\end{compactenum}
Mesela, a\c sa\u g\i daki tabloya bak\i n:
\begin{equation*}
\begin{array}{cc|cc|c}
P&Q&P\lto Q&P&Q\\\hline
0&0&1&0&0\\
1&0&0&1&0\\
0&1&1&0&1\\
1&1&1&1&1
\end{array}
\end{equation*}
Her sat\i rda,
\begin{compactitem}
  \item
ya $d(P\lto Q)=0$ veya $d(P)=0$,
\item
ya da $d(Q)=1$.
\end{compactitem}
  O zaman 
\begin{equation}\label{eqn:mp}
P\lto Q,P\models Q
\end{equation}
(yani, $\{P\lto Q,P\}\models Q$).  

Ayn\i\ \c sekilde,
\begin{equation}\label{eqn:ger}
P\lor Q\lor R,P\lto Q,Q\lto R\models R,
\end{equation}
\c c\"unk\"u, a\c sa\u g\i daki do\u gruluk tablosundaki her sat\i
rda,
\begin{compactitem}
  \item
ya $d(P\lor Q\lor R)=0$, veya $d(P\lto Q)=0$, veya $d(Q\lto R)=0$,
\item
ya da $R=1$.
\end{compactitem}
(\.Ikisi de olabilir, 6.\ sat\i rdaki gibi.)
\begin{equation*}
\begin{array}{ccc|ccc|c}
P&Q&R&P\lor Q\lor R&P\lto Q&Q\lto R&R\\\hline
0&0&0&0&1&1&0\\
1&0&0&1&0&1&0\\
0&1&0&1&1&0&0\\
1&1&0&1&1&0&0\\
0&0&1&1&1&1&1\\
1&0&1&1&0&1&1\\
0&1&1&1&1&1&1\\
1&1&1&1&1&1&1
\end{array}
\end{equation*}
Ama \c simdi
\begin{equation*}
\Gamma=\{\lnot(S\land T),(R\land Q)\lor(T\land Q),P\lor(S\land\lnot T), \lnot T\lor(Q\land(S\lor R)), \lnot R\lor T\}
\end{equation*}
olsun.  O zaman
\begin{equation}\label{eqn:Gamma}
\Gamma\models P\land Q\land R\land\lnot S\land T;
\end{equation}
ama bunu do\u gruluk tablolar\i yla g\"ostermek s\i k\i c\i\ olurdu.  \emph{Bi\c
cimsel kan\i t} y\"ontemi, bu durumda hem daha k\i sa, hem daha ilgin\c ctir. 

\textbf{Bi\c cimsel kan\i t,}
bir form\"uller listesidir.
 $F_0,F_1,\dots,F_n$, bir bi\c cimsel kan\i t olsun.
Her $k$ i\c cin, $0\leq k\leq n$ ise,
\begin{compactenum}
\item
ya $F_k$, bilinen bir totolojidir,
\item
ya bilinen bir gerektirmesine g\"ore,
$\{F_0,\dots,F_{k-1}\}\models F_k$,
\item
ya da $F_k$, bi\c cimsel kan\i t\i n bir \textbf{hipotezidir.}
\end{compactenum}
Kan\i t\i n \textbf{sonucu,} $F_n$ \"onerme form\"ul\"ud\"ur.  Kan\i t\i
n hipotezleri, $\Gamma$ k\"umesini olu\c stursun.  O zaman $\Gamma$, $F_n$ form\"ul\"un\"u gerektirir:
\begin{equation*}
  \Gamma\models F_n.
\end{equation*}
Bi\c cimsel kan\i t, $\Gamma$'dan $F_n$ form\"ul\"un\"u kan\i tlar.

Bilinen bir totolojiye \textbf{aksiyom} denir.  Ba\c slang\i\c cta,
aksiyom olarak, 
\begin{align*}
&1,&
&P\lor\lnot P
\end{align*}
form\"ullerini, ve daha genellikle her $F\lor\lnot F$ form\"ul\"un\"u, 
kullanabiliriz.  Ba\c ska bir form\"ul\"u aksiyom olarak kullanmak
istersek, kullanabiliriz, ama \"once onun totoloji oldu\u gunu ispatlamal\i
y\i z (mesela, do\u gruluk tablosuyla). 

Bilinen bir gerektirmeye,
\textbf{\c c\i kar\i m kural\i} denir. 
Mesela,
zaten~\eqref{eqn:mp} sat\i r\i ndan
\begin{equation*}
P\lto Q,P\models Q
\end{equation*}
gerektirmesini biliyoruz.  O zaman, \c c\i kar\i m kural\i\ olarak,
\textbf{ay\i rma kural\i m\i z}\footnote{Bu kural, Latincede
  \textbf{\emph{modus ponens,}} 
\.Ingilizcede,
  \textbf{detachment.}}  var, yani:  $P\lto Q$ ve $P$
form\"ullerinden, $Q$ \c c\i kar.  Daha genel olarak,
$F\lto G$ ve $F$ form\"ullerinden, $G$ \c c\i kar.  Hatta daha
genel olarak, $F\lto G$ ve $F$ form\"ulleri, $\Gamma$ k\"umesinin \"o\u
geleriyse,  $\Gamma$ k\"umesinden $G$ \c c\i kar. 

Ayr\i ca, $F\sim G$ ise, o zaman
\begin{equation*}
F\models G.
\end{equation*}
Teorem~\ref{thm:denk}'den, bilinen denkliklerimiz var.  O zaman \c su \c
c\i kar\i m kural\i m\i z var:  Teorem~\ref{thm:denk}'e g\"ore,
$F\sim G$ ise, ve $F$, $\Gamma$ k\"umesinin \"o\u gesiyse, o zaman
$G$, $\Gamma$'dan \c c\i kar. 

\"Orne\u gin, $P\land Q\models Q$, \c c\"unk\"u bi\c cimsel bir
kan\i t yazabiliriz:
\begin{align*}
  &{\sv P}\land {\sv Q},&&  
 1,&&  
 \lnot {\sv P}\lor 1,&&  
 \lnot {\sv P}\lor \lnot {\sv Q}\lor {\sv Q},&& 
 \lnot({\sv P}\land {\sv Q})\lor {\sv Q},&&  
 ({\sv P}\land {\sv Q})\lto {\sv Q},&& 
 {\sv Q}
\end{align*}
A\c c\i klamalar\i m\i z\i\ eklersek:
  \begin{center}
    \begin{tabular}{r c l}
(1) & ${\sv P}\land {\sv Q}$ & [hipotez]\\
(2) & $1$ & [aksiyom]\\
(3) & $\lnot {\sv P}\lor 1$ & [2. sat\i rdan, fazlal\i kla]\\
(4) & $\lnot {\sv P}\lor\lnot {\sv Q}\lor {\sv Q}$ & [3. sat\i rdan, fazlal\i kla]\\
(5) & $\lnot({\sv P}\land {\sv Q})\lor {\sv Q}$ & [4. sat\i rdan, De Morgan kural\i yla]\\
(6) & $({\sv P}\land {\sv Q})\lto {\sv Q}$ & [5. sat\i rdan]\\
(7) & ${\sv Q}$ & [1 \& 6 sat\i r\i ndan ay\i rma]
    \end{tabular}
  \end{center}
Ayr\i ca,
\begin{equation*}
P\lor Q\lor R,\lnot P,\lnot Q\models R,
\end{equation*}
\c c\"unk\"u bi\c cimsel bir kan\i t yazabiliriz:
\begin{align*}
&P\lor Q\lor R,&&\lnot P\lto Q\lor R,&&\lnot P,&&Q\lor R,&&\lnot Q\lto R,&&\lnot Q,&&R.
\end{align*}
A\c c\i klamalar\i m\i z\i\ eklersek:
  \begin{center}
    \begin{tabular}{r c l}
(1) & $P\lor Q\lor R$ & [hipotez]\\
(2) & $\lnot P\lto Q\lor R$ & [1.\ sat\i rdan]\\
(3) & $\lnot P$ & [hipotez]\\
(4) & $Q\lor R$ & [2.\ ve 3.\ sat\i rlardan, ay\i rma kural\i yla]\\
(5) & $\lnot Q\lto R$ & [4.\ sat\i rdan]\\
(6) & $\lnot Q$ & [hipotez]\\
(7) & $R$ & [5.\ ve 6.\ sat\i rlardan]
\end{tabular}
\end{center}

Her bi\c cimsel kan\i t\i n arkas\i nda, bir a\u ga\c c vard\i r.  Mesela,
son kan\i t\i n a\u gac\i\ \c s\"oyle: 
\begin{center}
\newcommand{\treebox}[1]{\Tr{\psframebox{#1}}}
\pstree[treemode=D]
       {\treebox{$R$}}
       {\treebox{$\lnot Q$}
        \pstree{\treebox{$\lnot Q\lto R$}}
               {\pstree{\treebox{$Q\lor R$}}
                       {\treebox{$\lnot P$}
                        \pstree{\treebox{$\lnot P\lto Q\lor R$}}
                               {\treebox{$P\lor Q\lor R$}}}}}
\end{center}
Bu bi\c cimsel kan\i t\i n s\i ralanmas\i n\i\ de\u gi\c stirebiliriz;
\"orne\u gin, \c s\"oyle yazabiliriz: 
\begin{align*}
&\lnot Q,&
&P\lor Q\lor R,&
&\lnot P,&
&\lnot P\lto Q\lor R,&
&Q\lor R,&
&\lnot Q\lto R,&
&R.
\end{align*}

Yeni \c c\i kar\i m kurallar\i\ a\c sa\u g\i daki teoremden gelir:

\begin{theorem}
  Bu gerektirmelerimiz var:
  \begin{compactenum}
    \item
(Basitle\c stirme:)
      \begin{align*}
	P\land Q&\models P,&
	P\land Q&\models Q.
      \end{align*}
\item
(Ekleme:)
  \begin{align*}
    	P&\models P\lor Q,&
	Q&\models P\lor Q.
  \end{align*}
\item
(Ba\u glama:)
  \begin{equation*}
	P,Q\models P\land Q.
  \end{equation*}
\item
(Ay\i rma:)
  \begin{align*}
	P,P\lto Q&\models Q,&
	P\lor Q,\lnot P&\models Q,\\
	\lnot Q,P\lto Q&\models\lnot P,&
	P\lor Q,\lnot Q&\models P.
  \end{align*}
(hipotetik tas\i m:)
  \begin{equation*}
	P\lto Q,Q\lto R\models P\lto R.
  \end{equation*}
\item
(Olumlu dilemma:)
  \begin{equation*}
	P\lto Q,R\lto S,P\lor R\models Q\lor S.
  \end{equation*}
  \end{compactenum}
\end{theorem}

\begin{proof}
Al\i\c st\i rma.
\end{proof}

\c Simdi yukar\i daki~\eqref{eqn:ger} gerektirmesininin bu bi\c cimsel
kan\i d\i\ var (her sat\i r\i n nedeni nedir?): 
\begin{align*}
&P\lor Q\lor R&&\text{[hipotez]}\\
&(P\lor Q)\lor R&&\\
&\lnot(P\lor Q)\lto R&&\\
&P\lto Q&&\text{[hipotez]}\\
&Q\lto R&&\text{[hipotez]}\\
&P\lto R&&\\
&P\lor Q\lto R\lor R&&\\
&P\lor Q\lto R&&\\
&(P\lor Q)\lor\lnot(P\lor Q)\lto R\lor R&&\\
&(P\lor Q)\lor\lnot(P\lor Q)&&\\
&R\lor R&&\\
&R.
\end{align*}

Ayr\i ca,
yukar\i daki~\eqref{eqn:Gamma} gerektirmesinin,
Tablo~\ref{tab:kan}'deki
bi\c cimsel
kan\i t\i\ var.
\begin{table}
\setlength{\columnseprule}{0.5pt}
\begin{multicols}2
  \begin{gather*}
    (R\land Q)\lor(T\land Q)\\
(R\lor T)\land Q\\
Q\\
R\lor T\\
\lnot R\lor T\\
(R\lor T)\land(\lnot R\lor T)\\
(R\land\lnot R)\lor T\\
1\lor T\\
T\\
\lnot(S\land T)\\
\lnot S\lor\lnot T\\
\lnot\lnot T\\
\lnot S
\end{gather*}

\begin{gather*}
\lnot S\land T\\
P\lor(S\land\lnot T)\\
\lnot S\lor\lnot\lnot T\\
\lnot(S\land\lnot T)\\
P\\
\lnot T\lor(Q\land(S\lor R))\\
Q\land(S\lor R)\\
S\lor R\\
R\\
R\land\lnot S\land T\\
Q\land R\land\lnot S\land T\\
P\land Q\land R\land\lnot S\land T
  \end{gather*}
  \end{multicols}
  \caption{Bi\c cimsel bir kan\i t}\label{tab:kan}
 \end{table}

\begin{exercise}
  A\c sa\u g\i daki gerektirmeler i\c cin bi\c cimsel kan\i
  tlar\i\ yaz\i n.
  \begin{enumerate}
  \item 
$\models P\lto P\lto P$
\item
$\models P\lto Q\lto P$
\item
$\models P\lor(P\lto Q)$
\item
$\models (P\lto Q)\lor\lnot Q$
\item
$P\lto Q\land R\models P\lto Q$
\item
$P\land\lnot P\models Q$
\item
$P\land(Q\lor R)\models P\liff(\lnot Q\lor P)$
\item
$P\lto Q,P\lto\lnot Q\models\lnot Q$
\item
$P\lto R,Q\lto R\models P\lor Q\lto R$
\item
$P\lto R,Q\lto S\models P\lor Q\lto R\lor S$
  \end{enumerate}
\end{exercise}

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%\bibliography{../references}

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\end{thebibliography}


\end{document}