%! 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}