Enderton (logic). Finish all but exercise 1.2.14.

finite-set-exercises
Joshua Potter 2023-08-21 18:30:16 -06:00
parent c458eca8ae
commit 6c5fbf825a
2 changed files with 270 additions and 25 deletions

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@ -118,6 +118,9 @@
a function $$v \colon \mathcal{S} \rightarrow \{F, T\}$$ assigning either
$T$ or $F$ to each symbol in $\mathcal{S}$.
\suitdivider
\noindent
Let $\bar{\mathcal{S}}$ be the set of \nameref{ref:well-formed-formula}s that
can be built up from $\mathcal{S}$ by the five
\nameref{ref:formula-building-operations}.
@ -246,6 +249,7 @@
{Enderton.Logic.Chapter\_1.Wff.rec}
\begin{proof}
We note every well-formed formula can be characterized by a
\nameref{ref:construction-sequence}.
For natural number $m$, let $P(m)$ be the statement:
@ -1586,7 +1590,7 @@
\end{proof}
\subsection{\sorry{Exercise 1.2.10}}%
\subsection{\unverified{Exercise 1.2.10}}%
\hyperlabel{sub:exercise-1.2.10}
Say that a set $\Sigma_1$ of wffs is \textit{equivalent} to a set $\Sigma_2$
@ -1596,36 +1600,126 @@
tautologically implied by the remaining members in $\Sigma$.
Show that the following hold.
\subsubsection{\sorry{Exercise 1.2.10a}}%
\subsubsection{\unverified{Exercise 1.2.10a}}%
\hyperlabel{ssub:exercise-1.2.10a}
A finite set of wffs has an independent equivalent subset.
\begin{proof}
TODO
For natural number $n$, let $P(n)$ be the statement:
\begin{induction}
\hyperlabel{sub:exercise-1.2.10a-ih}
A set of wffs \nameref{S:ref:equinumerous} to $n$ has an independent
equivalent subset.
\end{induction}
\noindent
We proceed by induction on $n$.
\paragraph{Base Case}%
Consider a finite set of wffs equinumerous to $0$.
This is simply the empty set.
It is vacuously true that $\emptyset$ is independent.
Thus $\emptyset \subseteq \emptyset$ is an independent equivalent subset
meaning $P(0)$ is true.
\paragraph{Inductive Step}%
Suppose $P(n)$ holds true for some $n \geq 0$.
That is, every finite set of wffs equinumerous to $n$ has an independent
equivalent subset.
Consider now set $\Sigma$ of wffs equinumerous to $n + 1$.
There are two possibilities to consider:
\subparagraph{Case 1}%
Suppose $\Sigma$ is independent.
Then $\Sigma \subseteq \Sigma$ is an independent equivalent subset.
\subparagraph{Case 2}%
Suppose $\Sigma$ is not independent.
Then there exists some $\sigma \in \Sigma$ such that $\sigma$ is
tautologically implied by the remaining members of $\Sigma$.
Let $\Sigma_1 = \Sigma - \{\sigma\}$.
By \ihref{sub:exercise-1.2.10a-ih}, $\Sigma_1$ has an independent
equivalent subset $\Sigma_2$.
Now let $\phi$ be an arbitrary wff.
Then
\begin{align*}
\Sigma_2 \vDash \phi
& \Rightarrow \Sigma_1 \vDash \phi
& \text{def'n of equivalent} \\
& \Rightarrow \Sigma_1; \sigma \vDash \phi
& \sigma \text{ is redundant} \\
& \Rightarrow \Sigma \vDash \phi.
\end{align*}
Likewise,
\begin{align*}
\Sigma \vDash \phi
& \Rightarrow \Sigma_1; \sigma \vDash \phi \\
& \Rightarrow \Sigma_1 \vDash \phi
& \sigma \text{ is redundant} \\
& \Rightarrow \Sigma_2 \vDash \phi.
& \text{def'n of equivalent}
\end{align*}
Thus $\Sigma_2$ is an independent equivalent subset of $\Sigma$.
\subparagraph{Subconclusion}%
The above two cases are exhaustive.
Hence $P(n + 1)$ holds true.
\paragraph{Conclusion}%
By induction, it follows $P(n)$ holds true for all $n \geq 0$.
That is, every set of wffs equinumerous to a natural number has an
independent equivalent subset.
In other words, every finite set of wffs has an independent equivalent
subset.
\end{proof}
\subsubsection{\sorry{Exercise 1.2.10b}}%
\subsubsection{\unverified{Exercise 1.2.10b}}%
\hyperlabel{ssub:exercise-1.2.10b}
An infinite set need not have an independent equivalent subset.
\begin{proof}
TODO
Let $$S = \{A_1 \land \cdots \land A_n \mid n \in \omega\}$$ be an infinite
set of wffs.
For the sake of contradiction, suppose $S$ has an independent equivalent
subset $S'$.
There are two cases to consider:
\paragraph{Case 1}%
Suppose $S' = \emptyset$.
Then it trivally follows $S'$ is not equivalent to $S$, a contradiction.
\paragraph{Case 1}%
Suppose $S' \neq \emptyset$.
By the \nameref{S:sub:well-ordering-natural-numbers}, there exists a least
$n \in \mathbb{N}$ such that $\phi = A_1 \land \cdots \land A_n$ is in
$S'$.
It cannot be that another element of $S'$ exists since such an element
would tautologically imply $\phi$, contradicting independence.
Thus $S' = \{\phi\}$.
But $\{\phi\}$ cannot be equivalent to $S$ since it has no information
about sentence symbol e.g. $A_{n+1}$, another contradiction.
\paragraph{Conclusion}%
The above two cases are exhaustive and both yield contradictions.
It must be that $S$ does not have an independent equivalent subset.
\end{proof}
\subsubsection{\sorry{Exercise 1.2.10c}}%
\hyperlabel{ssub:exercise-1.2.10c}
Let $\Sigma = \{\sigma_0, \sigma_1, \ldots\}$; show that there is an
independent equivalent set $\Sigma'$.
(By part (b), we cannot hope to have $\Sigma' \subseteq \Sigma$ in general.)
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 1.2.11}}%
\subsection{\unverified{Exercise 1.2.11}}%
\hyperlabel{sub:exercise-1.2.11}
Show that a truth assignment $v$ satisfies the wff
@ -1636,10 +1730,88 @@
parentheses is not crucial.)
\begin{proof}
TODO
Define $\sigma_n$ recursively as follows:
$\sigma_0 = (A_1 \Leftrightarrow A_2)$ and
$\sigma_{n+1} = (\sigma_n \Leftrightarrow A_{n+3})$.
For natural number $n$, let $P(n)$ be the statement:
\begin{induction}
\hyperlabel{sub:exercise-1.2.11-ih}
Truth assignment $v$ satisfies $\sigma_n$ if and only if $v(A_i) = F$
for an even number of $i$'s, $1 \leq i \leq n + 2$.
\end{induction}
\noindent
We proceed by induction on $n$.
\paragraph{Base Case}%
Let $n = 0$.
Then $\sigma_n = \sigma_0 = (A_1 \Leftrightarrow A_2)$.
We proceed by truth table:
$$\begin{array}{s|e|s}
(A_1 & \Leftrightarrow & A_2) \\
\hline
T & T & T \\
T & F & F \\
F & F & T \\
F & T & F
\end{array}$$
Here we see $A_1 \Leftrightarrow A_2$ is true if and only if both $A_1$
and $A_2$ are true or neither $A_1$ nor $A_2$ are true.
Thus $P(0)$ holds true.
\paragraph{Inductive Step}%
Suppose $P(n)$ holds true for some $n \geq 0$.
Consider now $$\sigma_{n+1} = (\sigma_k \Leftrightarrow A_{n+3}).$$
Let $v$ be a truth assignment for $A_1, \ldots, A_{n+3}$.
There are two cases to consider:
\subparagraph{Case 1}%
Suppose $v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 2$.
By \ihref{sub:exercise-1.2.11-ih}, $v$ satisfies $\sigma_n$.
We now have the following truth table:
$$\begin{array}{s|e|s}
(\sigma_n & \Leftrightarrow & A_{n+3}) \\
\hline
T & T & T \\
T & F & F \\
\end{array}$$
In this case, it follows $v$ satisfies $\sigma_{n+1}$ if and only if
$v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 3$.
\subparagraph{Case 2}%
Suppose $v(A_i) = F$ for an odd number of $i$'s, $1 \leq i \leq n + 2$.
By \ihref{sub:exercise-1.2.11-ih}, $v$ does not satisfy $\sigma_n$.
We now have the following truth table:
$$\begin{array}{s|e|s}
(\sigma_n & \Leftrightarrow & A_{n+3}) \\
\hline
F & F & T \\
F & T & F \\
\end{array}$$
In this case, it follows $v$ satisfies $\sigma_{n+1}$ if and only if
$v(A_i) = F$ for an even number of $i$'s, $1 \leq i \leq n + 3$.
\subparagraph{Subconclusion}%
The above two cases are exhaustive.
Hence $P(n + 1)$ holds true.
\paragraph{Conclusion}%
By induction, it follows $P(n)$ holds true for all $n \geq 0$.
That is, truth assignment $v$ satisfies
$$(\cdots (A_1 \Leftrightarrow A_2)
\Leftrightarrow \cdots \Leftrightarrow A_n)$$
if and only if $v(A_i) = F$ for an even number of $i$'s,
$1 \leq i \leq n$.
\end{proof}
\subsection{\sorry{Exercise 1.2.12}}%
\subsection{\unverified{Exercise 1.2.12}}%
\hyperlabel{sub:exercise-1.2.12}
There are three suspects for a murder: Adams, Brown, and Clark.
@ -1647,17 +1819,20 @@
But Clark hated him."
Brown states "I didn't do it. I didn't even know the guy. Besides I was out of
town all that week."
Clark says "I didn't do it. I saw both ADams and Brown downtown with the
Clark says "I didn't do it. I saw both Adams and Brown downtown with the
victim that day; one of them must have done it."
Assume that the two innocent men are telling the truth, but that the guilty
man might not be.
Who did it?
\begin{proof}
TODO
It must be that Brown is the guilty one.
Adam claims the victim was an old acquaintance of Brown's.
Clark claims Brown was downtown with the victim that day.
Brown's testimony conflicts with both of these statements.
\end{proof}
\subsection{\sorry{Exercise 1.2.13}}%
\subsection{\unverified{Exercise 1.2.13}}%
\hyperlabel{sub:exercise-1.2.13}
An advertisement for a tennis magazine states, "If I'm not playing tennis,
@ -1670,7 +1845,17 @@
truth assignments.)
\begin{proof}
TODO
Let $P$ denote playing tennis, $W$ denote watching tennis, and $R$ denote
reading about tennis.
These statements can be translated as:
\begin{enumerate}[(a)]
\item $\neg P \Rightarrow W$.
\item $\neg W \Rightarrow R$.
\end{enumerate}
Thus either the speaker is playing tennis, or, if not, he is watching
tennis.
Since we assume the speaker cannot do more than one of these activities at
a time, reading is never a possibility.
\end{proof}
\subsection{\sorry{Exercise 1.2.14}}%
@ -1684,10 +1869,11 @@
\nameref{sub:induction-principle-1} to show that $\bar{v}_1 = \bar{v}_2$.
\begin{proof}
The conditions 0-5 can be found at \nameref{ref:truth-assignment}.
TODO
\end{proof}
\subsection{\sorry{Exercise 1.2.15}}%
\subsection{\verified{Exercise 1.2.15}}%
\hyperlabel{sub:exercise-1.2.15}
Of the following three formulas, which tautologically implies which?
@ -1697,8 +1883,51 @@
\item $(((\neg A) \lor B) \land (A \lor (\neg B)))$
\end{enumerate}
\code{Bookshelf/Enderton/Logic/Chapter\_1}
{Enderton.Logic.Chapter\_1.exercise\_1\_2\_15\_i}
\code{Bookshelf/Enderton/Logic/Chapter\_1}
{Enderton.Logic.Chapter\_1.exercise\_1\_2\_15\_ii}
\begin{proof}
TODO
All three are tautologically equivalent.
We prove that (i) (a) is tautologically equivalent to (b) and (ii) (a) is
tautologically equivalent to (c). It then immediately follows that (b) is
tautologically equivalent to (c).
\paragraph{(i)}%
By \nameref{sub:exercise-1.2.4}, $(a) \vDash\Dashv (b)$ if and only if
$\vDash ((a) \Leftrightarrow (b))$.
We now construct the corresponding truth table:
$$\begin{array}{s|c|s|e|c|s|c|s|c|c|s|c|s}
(A & \Leftrightarrow & B) & \Leftrightarrow &
(\neg & ((A & \Rightarrow & B) &
\Rightarrow & (\neg & (B & \Rightarrow & A)))) \\
\hline
T & T & T & T & T & T & T & T & F & F & T & T & T \\
T & F & F & T & F & T & F & F & T & F & F & T & T \\
F & F & T & T & F & F & T & T & T & T & T & F & F \\
F & T & F & T & T & F & T & F & F & F & F & T & F
\end{array}$$
Therefore (a) and (b) are tautologically equivalent.
\paragraph{(ii)}%
By \nameref{sub:exercise-1.2.4}, $(a) \vDash\Dashv (c)$ if and only if
$\vDash ((a) \Leftrightarrow (c))$.
We now construct the corresponding truth table:
$$\begin{array}{s|c|s|e|c|s|c|s|c|s|c|c|s}
(A & \Leftrightarrow & B) & \Leftrightarrow &
(((\neg & A) & \lor & B) & \land & (A & \lor & (\neg & B))) \\
\hline
T & T & T & T & F & T & T & T & T & T & T & F & T \\
T & F & F & T & F & T & F & F & F & T & T & T & F \\
F & F & T & T & T & F & T & T & F & F & F & F & T \\
F & T & F & T & T & F & T & F & T & F & T & T & F
\end{array}$$
\end{proof}
\end{document}

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@ -519,4 +519,20 @@ theorem exercise_1_2_6b
: (False True) ∧ ¬ False := by
simp
/-! #### Exercise 1.2.15
Of the following three formulas, which tautologically implies which?
(a) `(A ↔ B)`
(b) `(¬((A → B) →(¬(B → A))))`
(c) `(((¬ A) B) ∧ (A (¬ B)))`
-/
theorem exercise_1_2_15_i (A B : Prop)
: (A ↔ B) ↔ (¬((A → B) → (¬(B → A)))) := by
tauto
theorem exercise_1_2_15_ii (A B : Prop)
: (A ↔ B) ↔ (((¬ A) B) ∧ (A (¬ B))) := by
tauto
end Enderton.Logic.Chapter_1