diff --git a/Bookshelf/Enderton/Logic.tex b/Bookshelf/Enderton/Logic.tex index e7de261..1132037 100644 --- a/Bookshelf/Enderton/Logic.tex +++ b/Bookshelf/Enderton/Logic.tex @@ -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} diff --git a/Bookshelf/Enderton/Logic/Chapter_1.lean b/Bookshelf/Enderton/Logic/Chapter_1.lean index 004c9ca..15210fd 100644 --- a/Bookshelf/Enderton/Logic/Chapter_1.lean +++ b/Bookshelf/Enderton/Logic/Chapter_1.lean @@ -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