%! TEX root = LST.tex
% vim: tw=50
% 18/01/2024 09AM

% Andras Zsak
% Outline:
% Chapter 1: Propositional Logic
% Chapter 2: Well-orderings and ordinals
% Chapter 3: Posets and Zorn's Lemma
% Chapter 4: Predicate Logic
% Chapter 5: Set Theory
% Chapter 6: Cardinal Arithmetic
% Chapter 7*: ?

% 4 Example Sheets
% Prerequisites: very little, countability, linear algebra,
% groups, fields

% Books:
%   1. PT. Johnstone: Notes on Logic and Set Theory
%   2. D. Von Dalen: Logic and Structure (good for Chapter 4)
%   3. A. Hajnal & P. Hamburger: Set Theory (good for ordinals and cardinals)
%   4. T. Jech: Set Theory (advanced)

\section{Propositional Logic}
\label{sec1}

We build a language consisting of statements /
propositions; we will assign truth values to
statements; we build a deduction system so that we
can prove statements that are true (and only
those).

These are also the features of more complicated
languages.

\begin{flashcard}[language-defn]
\begin{definition*}[Language of Propositional Logic]
  \glsnoundefn{primp}{primitive
  proposition}{primitive propositions}%
  \glsnoundefn{prop}{proposition}{propositions}%
  \glssymboldefn{false}{$\perp$}{$\perp$}
  \glssymboldefn{propL}{L}{L}
  \glssymboldefn{propP}{P}{P}
  \cloze{
  Our language consists of a set $P$ of
  \emph{primitive propositions} and a set $L =
  L(P)$ of \emph{propositions} defined inductively
  as follows:
  \begin{enumerate}[(i)]
    \item $P \subset L$
    \item $\false\, \in L$ ($\false$ is called
      `false' or `bottom')
    \item If $p, q \in L$ then $(p \impl q) \in
      L$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\vspace{-1em}
Often $P = \{p_1, p_2, p_3, \ldots\}$.

\begin{example*}
  $(p_1 \impl p_2)$, $((p_1 \impl \false)
  \impl p_2)$, $((p_1 \impl p_2) \impl
  (p_1 \impl p_3))$. If $p \in \propL$ then we must
  always have $((p \impl \false) \impl
  \false) \in \propL$.
\end{example*}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \glssymboldefn{propLnum}{$L_1$}{$L_1$}
      ``Defined inductively'' means that
      $\propL =
      \bigcup_{n \in \NN} L_n$ where
      \begin{align*}
        L_1
        &= P \cup \{\false\} \\
        L_{n + 1}
        &= L_n \cup \{(p \impl q) \st p, q \in
        L_n\} \qquad n \in \NN
      \end{align*}
    \item Every $p \in \propL$ is a finite string
      in $\propP \cup \{\false, \impl, (, )\}$.
      Can prove that $L$ is the smallest (with
      respect to inclusion) subset of the set
      $\Sigma$ of all finite strings in $\propP \cup
      \{\false, \impl, (, )\}$ such that (i) -
      (iii) above hold. Note $\propL \subsetneq
      \Sigma$. For example, $\impl p_1 p_3( \in
      \Sigma \setminus \propL$.
    \item Every $p \in \propL$ is uniquely
      determined by (i) - (iii) above, i.e. either
      $p \in \propP$ or $p = \,\false$ or there
      exists unique $q, r \in \propL$ such that $p
      = (q \impl r)$.
  \end{enumerate}
\end{remark*}

\glssymboldefn{and}{$\wedge$}{$\wedge$}%
\glssymboldefn{or}{$\vee$}{$\vee$}%
\glssymboldefn{not}{$\neg$}{$\neg$}%
\glssymboldefn{true}{$\top$}{$\top$}%

\vspace{-1em}
What about $\pand$, $\por$ etc? We introduct
symbols $\pand$ (`and'), $\por$ (`or'), $\true$
(`true' or `top') and $\pnot$ (`not') as
abbreviations as follows:
\begin{itemize}
  \item $\true = (\false \impl \false)$
  \item $\pnot p = (p \impl \false)$
  \item $p \por q = (\pnot p \impl q)$
  \item $p \pand q = \pnot(p \impl \pnot q)$
\end{itemize}

\subsection{Semantic Entailment}

\begin{flashcard}[valuation-defn]
\begin{definition*}[Valutation]
  \glsnoundefn{vtion}{valuation}{valuations}%
  \glssymboldefn{propv}{v}{v}
  \cloze{
  A \emph{valuation} on $\propL$ is a function $v
  : \propL \to \{0, 1\}$ such that
  \begin{enumerate}[(i)]
    \item $v(\false) = 0$
    \item if $p, q \in \propL$ then
      \[ v(p \impl q) = \begin{cases}
        0 & \text{if $v(p) = 1$ and $v(q) = 0$} \\
        1 & \text{otherwise}
      \end{cases} \]
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  $\propv(p_1) = 1$, $\propv(p_2) = 0$. Then
  \[ \propv(\ub{(\false \impl p_1)}_{1}
  \impl \ub{(p_1 \impl p_2)}_{0}) = 0 .\]
\end{example*}

\begin{flashcard}[vtion-uniqueness-prop]
\begin{proposition}
  \phantom{}
  \begin{enumerate}[(i)]
    \item If $\propv, \propv'$ are \glspl{vtion}
      on $\propL$ and $\propv|_{\propP} =
      \propv'|_{\propP}$ then $\propv = \propv'$.
    \item For any $w : \propP \to \{0, 1\}$, there
      is a \gls{vtion} $\propv : \propL \to \{0,
      1\}$ such that $\propv|_{\propP} = w$.
  \end{enumerate}
\end{proposition}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{So $\propv(p) = \propv'(p) ~\forall p
      \in \propP$ and $\propv(\false) =
      \propv'(\false) = 0$, so
      $\propv|_{\propLnum{1}} =
      \propv'|_{\propLnum{1}}$. If
      $\propv|_{\propLnum{n}} =
      \propv'|_{\propLnum{n}}$ then $\forall p, q
      \in \propLnum{n}$, $\propv(p \impl q) =
      \propv'(p \impl q)$ and thus
      $\propv|_{\propLnum{n + 1}} =
      \propv'|_{\propLnum{n + 1}}$. So by
      induction, $\propv$ and $\propv'$ agree on
      $\bigcup_n \propLnum{n} = \propL$.}
    \item \cloze{We define $\propv$ on $\propLnum{n}$ by
      induction: Let $\propv(p) = w(p) ~\forall p
      \in \propP$ and $\propv(\false) = 0$. This
      defines $\propv$ on $\propLnum{1}$. Assume
      $v$ is defined on $\propLnum{n}$. Given $p
      \in \propLnum{n + 1} \setminus
      \propLnum{n}$, write $p = (q \impl r)$,
      $q, r \in \propLnum{n}$ and define
      \[ \propv(p) = \begin{cases}
        0 & \text{if $\propv(q) = 1$, $\propv(r) =
        0$} \\
        1 & \text{otherwise}
      \end{cases} \]
      This defines $\propv$ on $\propLnum{n + 1}$.
      Hence $\propv$ is defined on $\bigcup_n
      \propLnum{n} = \propL$. By construction,
      $\propv$ is a \gls{vtion} on $\propL$ and
      $\propv|_{\propP} = w$. \qedhere}
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[tautology-defn]
\begin{definition*}[Tautology]
  \glsnoundefn{taut}{tautology}{tautologies}
  \cloze{$t \in L$ is a \emph{tautology} if
  $\propv(t) = 1$ for all \glspl{vtion} $v$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $(p \impl (q \impl p))$, $p, q \in
      \propL$ (a true statement is implied by any
      statement). We check:
      \begin{center}
        \begin{tabular}{c|c|c|c}
          $\propv(p)$ & $\propv(q)$
          & $\propv(q \impl p)$
          & $\propv(p \impl (q \impl p))$ \\
          \hline
          $0$ & $0$ & $1$ & $1$ \\
          $1$ & $0$ & $1$ & $1$ \\
          $0$ & $1$ & $0$ & $1$ \\
          $1$ & $1$ & $1$ & $1$
        \end{tabular}
      \end{center}

      \vspace{-3em}
