intro_to_logic.tex

\documentclass{article}
\title{Reference sheet for ``Introduction to Logic''}
\date{2019-04-15}
\author{Michael Lazear}
\pagestyle{headings}
\usepackage{amsmath}

\begin{document}
	\pagenumbering{gobble}
	\maketitle
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\tableofcontents
\newpage
\section{Introduction}

This document is a reference sheet for the book ``Introduction to Logic'' by Harry Gensler and This document \footnote{http://www.math.uvic.ca/faculty/gmacgill/guide/Logic.pdf}

\newpage
\section{Definitions and terms}
Definitions from an ``Introduction to Logic''

\medskip
An \textbf{argument} is a set of statements consisting of premises and a conclusion. An argument is \textbf{valid} if it is contradictory to have the premises all true and the conclusion false. -- $(p \wedge p_1 \wedge p_2 \wedge \ldots p_n) \rightarrow q$.
An argument is \textbf{sound} if it is valid and all of the premises are true.

\medskip
\textbf{wff}, a well-formed formula

\medskip
A statement that is  true in all cases is a \textbf{tautology} --  $(p \lor \neg p)$

\medskip
A \textbf{contradiction} is a statement that is false in all cases -- $(p \wedge \neg p)$

\medskip
\textbf{law of the excluded middle} states that every statement is true or false.

\newpage
\section{Propositional Logic}

\subsection{Compound statements}

The \textbf{conjuction of $P$ and $Q$} (e.g. \textbf{$P$ and $Q$}) $P \wedge Q$, states that both $P$ and $Q$ are true.

\begin{tabular}{cc|c}
P & Q & $(P \wedge Q)$ \\
\hline
0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\end{tabular}

\bigskip
The \textbf{disjunction of $p$ and $q$} (e.g. \textbf{$p$ or $q$}) $p \lor q$, states that either $p$ is true, or $q$ is true, or both are true (at least one part is true).

\begin{tabular}{cc|c}
P & Q & $(P \lor Q)$ \\
\hline
0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1 \\
\end{tabular}

\bigskip
The \textbf{conditional $p \rightarrow q$} (e.g. \textbf{if $p$ then $q$}), states that $q$ is not false if $p$ is true.

\begin{tabular}{cc|c}
P & Q & $(P \rightarrow Q)$ \\
\hline
0 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\end{tabular}

\bigskip
The \textbf{biconditional $p \leftrightarrow q$} (e.g. \textbf{$p$ if and only if $q$}), states that $p$ and $q$ have the same truth value.

\begin{tabular}{cc|c}
P & Q & $(P \leftrightarrow Q)$ \\
\hline
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
\end{tabular}

\bigskip
The \textbf{negation of $p$} states that if $p$ is true, then $\neg p$ is false, and if $p$ is false, then $\neg p$ is true.

\begin{tabular}{c|c}
P & $\neg P$ \\
\hline
0 & 1 \\
1 & 0 \\
\end{tabular}

\bigskip
\begin{tabular}{cc|cccc}
P & Q & $(P \wedge Q)$ & $(P \lor Q)$ & $(P \rightarrow Q)$ & $(P \leftrightarrow Q)$ \\
\hline
0 & 0 & 0 & 0 & 1 & 1\\
0 & 1 & 0 & 1 & 1 & 0\\
1 & 0 & 0 & 1 & 0 & 0\\
1 & 1 & 1 & 1 & 1 & 1\\
\end{tabular}

\newpage
\subsection{S-rules, simplifications}
Some logical connectives can be simplified without further information.

\medskip
\textbf{And} \ldots both $p$ and $q$ are true
\begin{equation}
\frac{(P \wedge Q)}{P, Q}
\end{equation}

\medskip
\textbf{Not either} \ldots both $p$ and $q$ are false
\begin{equation}
\frac{\neg (P \lor Q)}{\neg P, \neg Q}
\end{equation}

\medskip
\textbf{False implication} \ldots $p$ is true and $q$ is false
\begin{equation}
\frac{\neg (P \rightarrow Q)}{ P, \neg Q}
\end{equation}

\subsection{I-rules, inference}
\textbf{I-rules} are used to infer a conclusion from two premises

With a \textbf{not both} or \textbf{nand} statement, only one part is true, so affirmation of one part allows us to deny the other:
\begin{equation}
\begin{tabular}{c}
$\neg(P \wedge Q)$\\
$Q$ \\
\hline
$\neg P$\\
\end{tabular}
\tag{I-1}\label{eq:I-1}
\end{equation}
\begin{equation}
\begin{tabular}{c}
$\neg(P \wedge Q)$\\
$P$ \\
\hline
$\neg Q$\\
\end{tabular}
\tag{I-2}\label{eq:I-2}
\end{equation}
However we cannot make an inference given the negation of one of the statements:
\begin{tabular}{c}
$\neg(P \wedge Q)$ \\
$\neg P$ \\
\hline
nil\\
\end{tabular}


\bigskip
Denying one part of an \textbf{or} allows us to infer the truth value of the other part: 
\begin{equation}
\begin{tabular}{c}
$\neg(P \lor Q)$ \\
$\neg P$ \\
\hline
$Q$\\
\end{tabular}
\tag{I-3}\label{eq:I-3}
\end{equation}

\begin{equation}
\begin{tabular}{c}
$\neg(P \lor Q)$ \\
$\neg Q$ \\
\hline
$P$\\
\end{tabular}
\tag{I-4}\label{eq:I-4}
\end{equation}

\begin{equation}
(P \lor Q), \neg P \rightarrow Q
\end{equation}

\begin{equation}
(P \lor Q), \neg Q \rightarrow P
\end{equation}


\end{document}