r
/
bookshelf
1
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Enderton (logic). Begin proving inducting statements.

finite-set-exercises
Joshua Potter 2023-08-13 12:39:50 -06:00
parent bbae7136b1
commit 70e51584a5
3 changed files with 106 additions and 34 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

114
Bookshelf/Enderton/Logic.tex
View File

@ -18,9 +18,9 @@
\section{\defined{Construction Sequence}}%
\hyperlabel{ref:construction-sequence}
A \textbf{construction sequence} is a finite sequence
$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
for each $i \leq n$ we have at least one of
A \textbf{construction sequence} is a \nameref{ref:finite-sequence}
$\ltuple{\epsilon_1}{\epsilon_n}$ of \nameref{ref:expression}s such that for
each $i \leq n$ we have at least one of
\begin{align*}
& \epsilon_i \text{ is a sentence symbol} \\
& \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
@ -33,7 +33,14 @@
\section{\defined{Expression}}%
\hyperlabel{ref:expression}
An \textbf{expression} is a finite sequence of symbols.
An \textbf{expression} is a \nameref{ref:finite-sequence} of symbols.
\section{\defined{Finite Sequence}}%
\hyperlabel{ref:finite-sequence}
$S$ is a \textbf{finite sequence} (or \textbf{string}) of members of set $A$
if and only if, for some positive integer $n$, we have
$S = \ltuple{x_1}{x_n}$, where each $x_i \in A$.
\section{\defined{\texorpdfstring{$n$}{n}-tuple}}%
\hyperlabel{ref:n-tuple}
@ -46,10 +53,10 @@ We also define $\tuple{x} = x$.
\section{\defined{Well-Formed Formula}}%
\hyperlabel{ref:well-formed-formula}
An \nameref{ref:expression} that can be built up from the sentence symbols by
applying some finite number of times the
\textbf{formula-building operations} (on expressions) defined by the
equations:
A \textbf{well-formed formula} (wff) is an \nameref{ref:expression} that can
be built up from the sentence symbols by applying some finite number of
times the \textbf{formula-building operations} (on expressions) defined by
the equations:
\begin{align*}
\mathcal{E}_{\neg}(\alpha)
& = (\neg \alpha) \\
@ -82,11 +89,11 @@ We also define $\tuple{x} = x$.
\begin{proof}
For natural number $m$, let $P(m)$ be the statement:
\begin{assumption}
\hyperlabel{sec:lemma-0a-eq1}
\text{If } \ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}
\text{ then } x_1 = \ltuple{y_1}{y_{k+1}}.
\end{assumption}
\begin{induction}
\hyperlabel{sec:lemma-0a-ih}
If $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$
then $x_1 = \ltuple{y_1}{y_{k+1}}$.
\end{induction}
\noindent
We proceed by induction on $m$.
@ -101,16 +108,16 @@ We also define $\tuple{x} = x$.
Suppose for $m \geq 1$ that $P(m)$ is true and assume
\begin{equation}
\hyperlabel{sec:lemma-0a-eq2}
\hyperlabel{sec:lemma-0a-eq1}
\ltuple{x_1}{x_{m+1}} = \ltuple{y_1, \ldots, y_{m+1}}{y_{m+1+k}}.
\end{equation}
By definition of an \nameref{ref:n-tuple}, we can decompose
\eqref{sec:lemma-0a-eq2} into the following two identities
\eqref{sec:lemma-0a-eq1} into the following two identities
\begin{align*}
x_{m+1} & = y_{m+1+k} \\
\ltuple{x_1}{x_m} & = \ltuple{y_1}{y_{m+k}}.
\end{align*}
By \eqref{sec:lemma-0a-eq1}, $P(m)$ implies $x_1 = \ltuple{y_1}{y_{k+1}}$.
By \ihref{sec:lemma-0a-ih}, $P(m)$ implies $x_1 = \ltuple{y_1}{y_{k+1}}$.
Hence $P(m+1)$ holds true.
\paragraph{Conclusion}%
@ -125,7 +132,7 @@ We also define $\tuple{x} = x$.
\section{The Language of Sentential Logic}%
\hyperlabel{sec:language-sentential-logic}
\subsection{\sorry{Induction Principle}}%
\subsection{\unverified{Induction Principle}}%
\hyperlabel{sub:induction-principle-1}
\begin{theorem}
@ -135,7 +142,78 @@ We also define $\tuple{x} = x$.
\end{theorem}
\begin{proof}
TODO
We note every \nameref{ref:well-formed-formula} can be characterized by a
\nameref{ref:construction-sequence}.
For natural number $m$, let $P(m)$ be the statement:
\begin{induction}
\hyperlabel{sub:induction-principle-1-ih}
Every wff characterized by a construction sequence of length $m$ is in
$S$.
\end{induction}
\noindent
We proceed by strong induction on $m$.
\paragraph{Base Case}%
Let $\phi$ denote a wff characterized by a construction sequence of length
$1$.
Then it must be that $\phi$ is a single sentence symbol.
By hypothesis, $S$ contains all the sentence symbols.
Thus $P(1)$ holds true.
\paragraph{Inductive Step}%
Suppose $P(0)$, $P(1)$, $\ldots$, $P(m)$ holds true and let $\phi$ denote
a wff characterized by a construction sequence of length $m + 1$.
By definition of a construction sequence, one of the following holds:
\begin{align}
& \phi \text{ is a sentence symbol}
& \label{sub:induction-principle-1-eq1} \\
& \phi = \mathcal{E}_\neg(\epsilon_j)
\text{ for some } j < m + 1
& \label{sub:induction-principle-1-eq2} \\
& \phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
\text{ for some } j < m + 1, k < m + 1
& \label{sub:induction-principle-1-eq3}
\end{align}
where $\square$ is one of the binary connectives $\land$, $\lor$,
$\Rightarrow$, $\Leftrightarrow$.
We consider each case in turn.
\subparagraph{\eqref{sub:induction-principle-1-eq1}}%
By hypothesis, all sentence symbols are in $S$.
Thus $\phi \in S$.
\subparagraph{\eqref{sub:induction-principle-1-eq2}}%
Suppose $\phi = \mathcal{E}_\neg(\epsilon_j)$ for some $j < m + 1$.
By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ is in $S$.
By hypothesis, $S$ is closed under $\mathcal{E}_\neg$.
Thus $\phi \in S$.
\subparagraph{\eqref{sub:induction-principle-1-eq3}}%
Suppose $\phi = \mathcal{E}_\square(\epsilon_j, \epsilon_k)$ for some
$j < m + 1, k < m + 1$,
By \ihref{sub:induction-principle-1-ih}, $\epsilon_j$ and $\epsilon_k$
is in $S$.
By hypothesis, $S$ is closed under $\mathcal{E}_\square$ for all
possible candidates of $\square$.
Thus $\phi \in S$.
\subparagraph{Subconclusion}%
Since the above three cases are exhaustive, $P(m + 1)$ holds.
\paragraph{Conclusion}%
By strong induction, $P(m)$ holds true for all natural numbers $m \geq 1$.
Since every well-formed formula is characterized by a construction
sequence, the set of all wffs is a subset of $S$.
Likewise, it obviously holds that $S$ is a subset of all wffs.
Thus $S$ is precisely the set of all wffs.
\end{proof}
\section{Exercises 1}%

4
Bookshelf/Enderton/Set.tex
View File

@ -5929,7 +5929,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and
Thus $F(v_1(n)) = F(y) = F(v_2(n))$ and $h(n^+) = F(y)$.
Therefore $n^+ \in S$.
\subparagraph{Conclusion}%
\subparagraph{Subconclusion}%
By \nameref{spar:recursion-theorem-natural-numbers-i-1} and
\nameref{spar:recursion-theorem-natural-numbers-i-2}, $S$ is an
@ -6002,7 +6002,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and
\end{enumerate}
Since $v'$ is acceptable, $n^+ \in \dom{h}$.
\subparagraph{Conclusion}%
\subparagraph{Subconclusion}%
By \nameref{spar:recursion-theorem-natural-numbers-iii-1} and
\nameref{spar:recursion-theorem-natural-numbers-iii-2},

22
preamble.tex
View File

@ -92,21 +92,15 @@
% Admonitions
% ========================================
\newcommand\@assumptionbody[1]{
\begin{equation}
\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
#1
\end{equation}}
\NewEnviron{induction}[1][]{%
\def\title{\ifstrempty{#1}
{Induction Hypothesis (IH)}
{#1}}
\begin{tcolorbox}[title=\title]
\BODY
\end{tcolorbox}}
\NewEnviron{assumption}[1][]{%
\ifstrempty{#1}{%
\begin{tcolorbox}[bottom=8pt]
\@assumptionbody{\BODY}
\end{tcolorbox}}{%
\begin{tcolorbox}[title=#1,bottom=8pt]
\@assumptionbody{\BODY}
\end{tcolorbox}}}
\newcommand{\ihref}[1]{\hyperref[#1]{(IH)}}
\NewEnviron{note}{%
\begin{tcolorbox}[%