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\title{First theorems of Propositional Calculus}
\author{Michael Meyling}
\date{\tt <module@qedeq.org>}


\begin{document}

\maketitle

\setlongtables



{\small
This document is part of the project ``Hilbert II''.
To get more information about this project look at:\\
\mbox{\url{http://www.qedeq.org}}.


Permission is granted to copy, distribute and/or modify this document
under the terms of the GNU Free Documentation License, Version 1.2
or any later version published by the Free Software Foundation;
with no Invariant Sections, no Front-Cover Texts, and no
Back-Cover Texts. See also under \url{http://www.gnu.org/copyleft/}
}


\section*{Abstract}

This module includes first proofs of propositional calculus theorems.


\section*{Specification}

This document has the following specification:

\begin{longtable}[h!]{l@{\extracolsep{\fill}}p{12cm}}
Name: & proptheo1 \\
Version: & 1.00.00 \\
Rule version: & 1.00.00 \\
Orgin: & \url{http://www.qedeq.org/0_00_53/proptheo1_1.00.00_1.00.00.qedeq} \\
\end{longtable}

\medskip

Author of this module: 
\begin{longtable}[h!]{l@{\extracolsep{\fill}}l}
Michael Meyling & mime@qedeq.org \\
\end{longtable}




\section*{References}

This document uses the results of the following documents:

\begin{longtable}[h!]{l@{\extracolsep{\fill}}p{12cm}}
Name: & propaxiom \\
Version: & 1.00.00 \\
Rule version: & 1.00.00 \\
Orgin: & \url{propaxiom_1.00.00_1.00.00.qedeq}  \\
pdf: & \url{propaxiom_1.00.00_1.00.00.pdf}  \\
\end{longtable}


\section*{Content}




First we prove a well known tautology:




\begin{thm}[theo1]
\hypertarget{theo1}{}
\begin{displaymath}
\neg P\ \vee \  P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo1:1}
  $1$ & $(P\ \rightarrow \  Q)\ \rightarrow \  ((A\ \vee \  P)\ \rightarrow \  (A\ \vee \  Q))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom4}{axiom4} } \\
\label{theo1:2}
  $2$ & $((P\ \vee \  P)\ \rightarrow \  Q)\ \rightarrow \  ((A\ \vee \  (P\ \vee \  P))\ \rightarrow \  (A\ \vee \  Q))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $P$ by $P\ \vee \  P$ in \hyperref[theo1:1]{$1$}
} \\
\label{theo1:3}
  $3$ & $((P\ \vee \  P)\ \rightarrow \  P)\ \rightarrow \  ((A\ \vee \  (P\ \vee \  P))\ \rightarrow \  (A\ \vee \  P))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $Q$ by $P$ in \hyperref[theo1:2]{$2$}
} \\
\label{theo1:4}
  $4$ & $((P\ \vee \  P)\ \rightarrow \  P)\ \rightarrow \  ((\neg P\ \vee \  (P\ \vee \  P))\ \rightarrow \  (\neg P\ \vee \  P))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $A$ by $\neg P$ in \hyperref[theo1:3]{$3$}
} \\
\label{theo1:5}
  $5$ & $(P\ \vee \  P)\ \rightarrow \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom1}{axiom1} } \\
\label{theo1:6}
  $6$ & $(\neg P\ \vee \  (P\ \vee \  P))\ \rightarrow \  (\neg P\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo1:5]{$5$}, \hyperref[theo1:4]{$4$}} \\
\label{theo1:7}
  $7$ & $(P\ \rightarrow \  (P\ \vee \  P))\ \rightarrow \  (\neg P\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule6}{reverse abbreviation} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{impl}{impl} in \hyperref[theo1:6]{$6$} at occurence $1$
} \\
\label{theo1:8}
  $8$ & $P\ \rightarrow \  (P\ \vee \  Q)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom2}{axiom2} } \\
\label{theo1:9}
  $9$ & $P\ \rightarrow \  (P\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $Q$ by $P$ in \hyperref[theo1:8]{$8$}
} \\
\label{theo1:10}
  $10$ & $\neg P\ \vee \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo1:9]{$9$}, \hyperref[theo1:7]{$7$}} \\
 & & \qedhere
\end{longtable}
\end{proof}



We just use our first sentence to get the second theorem:




\begin{thm}[theo2]
\hypertarget{theo2}{}
\begin{displaymath}
P\ \rightarrow \  P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo2:1}
  $1$ & $\neg P\ \vee \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo1}{theo1} } \\
\label{theo2:2}
  $2$ & $P\ \rightarrow \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule6}{reverse abbreviation} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{impl}{impl} in \hyperref[theo2:1]{$1$} at occurence $1$
} \\
 & & \qedhere
\end{longtable}
\end{proof}



And another use of the first theorem, to get the law of the excluded middle (tertium non datur):




\begin{thm}[theo3]
\hypertarget{theo3}{}
\begin{displaymath}
P\ \vee \  \neg P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo3:1}
  $1$ & $\neg P\ \vee \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo1}{theo1} } \\
\label{theo3:2}
  $2$ & $(P\ \vee \  Q)\ \rightarrow \  (Q\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom3}{axiom3} } \\
\label{theo3:3}
  $3$ & $(\neg P\ \vee \  Q)\ \rightarrow \  (Q\ \vee \  \neg P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $P$ by $\neg P$ in \hyperref[theo3:2]{$2$}
} \\
\label{theo3:4}
  $4$ & $(\neg P\ \vee \  P)\ \rightarrow \  (P\ \vee \  \neg P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $Q$ by $P$ in \hyperref[theo3:3]{$3$}
} \\
\label{theo3:5}
  $5$ & $P\ \vee \  \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo3:1]{$1$}, \hyperref[theo3:4]{$4$}} \\
 & & \qedhere
\end{longtable}
\end{proof}



Also trivial is:




\begin{thm}[theo4]
\hypertarget{theo4}{}
\begin{displaymath}
P\ \rightarrow \  \neg \neg P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo4:1}
  $1$ & $P\ \vee \  \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo3}{theo3} } \\
\label{theo4:2}
  $2$ & $\neg P\ \vee \  \neg \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $P$ by $\neg P$ in \hyperref[theo4:1]{$1$}
} \\
\label{theo4:3}
  $3$ & $P\ \rightarrow \  \neg \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule6}{reverse abbreviation} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{impl}{impl} in \hyperref[theo4:2]{$2$} at occurence $1$
} \\
 & & \qedhere
\end{longtable}
\end{proof}



Three negations:




\begin{thm}[theo5]
\hypertarget{theo5}{}
\begin{displaymath}
P\ \vee \  \neg \neg \neg P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo5:1}
  $1$ & $P\ \rightarrow \  \neg \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo4}{theo4} } \\
\label{theo5:2}
  $2$ & $(P\ \rightarrow \  Q)\ \rightarrow \  ((A\ \vee \  P)\ \rightarrow \  (A\ \vee \  Q))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom4}{axiom4} } \\
\label{theo5:3}
  $3$ & $(P\ \rightarrow \  \neg \neg P)\ \rightarrow \  ((A\ \vee \  P)\ \rightarrow \  (A\ \vee \  \neg \neg P))$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $Q$ by $\neg \neg P$ in \hyperref[theo5:2]{$2$}
} \\
\label{theo5:4}
  $4$ & $(A\ \vee \  P)\ \rightarrow \  (A\ \vee \  \neg \neg P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo5:1]{$1$}, \hyperref[theo5:3]{$3$}} \\
\label{theo5:5}
  $5$ & $(A\ \vee \  \neg P)\ \rightarrow \  (A\ \vee \  \neg \neg \neg P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $P$ by $\neg P$ in \hyperref[theo5:4]{$4$}
} \\
\label{theo5:6}
  $6$ & $(P\ \vee \  \neg P)\ \rightarrow \  (P\ \vee \  \neg \neg \neg P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $A$ by $P$ in \hyperref[theo5:5]{$5$}
} \\
\label{theo5:7}
  $7$ & $P\ \vee \  \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo3}{theo3} } \\
\label{theo5:8}
  $8$ & $P\ \vee \  \neg \neg \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo5:7]{$7$}, \hyperref[theo5:6]{$6$}} \\
 & & \qedhere
\end{longtable}
\end{proof}



Now we could prove the reverse of Proposition 4:




\begin{thm}[theo6]
\hypertarget{theo6}{}
\begin{displaymath}
\neg \neg P\ \rightarrow \  P\end{displaymath}
\end{thm}
\begin{proof}
\mbox{}\\
\begin{longtable}[h!]{r@{\extracolsep{\fill}}p{9cm}@{\extracolsep{\fill}}p{4cm}}
\label{theo6:1}
  $1$ & $P\ \vee \  \neg \neg \neg P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule3}{add sentence} \hyperref{}{}{theo5}{theo5} } \\
\label{theo6:2}
  $2$ & $(P\ \vee \  Q)\ \rightarrow \  (Q\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule2}{add axiom} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{axiom3}{axiom3} } \\
\label{theo6:3}
  $3$ & $(P\ \vee \  \neg \neg \neg P)\ \rightarrow \  (\neg \neg \neg P\ \vee \  P)$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule4}{replace}  $Q$ by $\neg \neg \neg P$ in \hyperref[theo6:2]{$2$}
} \\
\label{theo6:4}
  $4$ & $\neg \neg \neg P\ \vee \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule1}{MP} with \hyperref[theo6:1]{$1$}, \hyperref[theo6:3]{$3$}} \\
\label{theo6:5}
  $5$ & $\neg \neg P\ \rightarrow \  P$
  & {\tiny \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{rule6}{reverse abbreviation} \hyperref{propaxiom_1.00.00_1.00.00.pdf}{}{impl}{impl} in \hyperref[theo6:4]{$4$} at occurence $1$
} \\
 & & \qedhere
\end{longtable}
\end{proof}


\end{document}