Merge pull request #30 from tenick/solve/5.4_exercise24.c

add proof for 5.4.24.c, formatting and minor fixes

Author: spamegg1

src/Epp.tex

@@ -2236,7 +2236,8 @@ \subsubsection{Exercise 12}
 \hline T & F & F & F \\
 \hline F & T & T & T \\
 \hline F & F & T & F \\
-\hline \end{array}
+\hline 
+\end{array}
 $$
 \end{proof}
 
@@ -2417,7 +2418,8 @@ \subsubsection{Exercise 20}
 \hline
 F & F & F & F \\
 \hline
-\end{array} $$
+\end{array}
+$$
 
 $p \wedge \false$ and $p \vee \false$ are not logically equivalent.
 \end{proof}
@@ -2863,7 +2865,8 @@ \subsubsection{Exercise 45}
 
 (b) It is not the case that both Bob and Ann are both math and computer science majors, but it is the case that Ann is a math major and Bob is both a math and computer science major.
 
-\begin{proof} Define
+\begin{proof}
+Define
 \begin{enumerate}
 \item $p$: Bob is a math major.
 \item $q$: Bob is a CS major.
@@ -2990,7 +2993,8 @@ \subsubsection{Exercise 46}
 \subsubsection{Exercise 47}
 In logic and in standard English, a double negative is equivalent to a positive.There is one fairly common English usage in which a “double positive” is equivalent to a negative. What is it? Can you think of others?
 
-\begin{proof} There is a famous story about a philosopher who once gave a talk in which he observed that whereas in English and many other languages a double negative is equivalent to a positive, there is no language in which a double positive is equivalent to a negative. To this, another philosopher, Sidney
+\begin{proof}
+There is a famous story about a philosopher who once gave a talk in which he observed that whereas in English and many other languages a double negative is equivalent to a positive, there is no language in which a double positive is equivalent to a negative. To this, another philosopher, Sidney
 Morgenbesser, responded sarcastically, “Yeah, yeah.”
 
 {\it [Strictly speaking, sarcasm functions like negation. When spoken sarcastically, the words “Yeah, yeah” are not a true double positive; they just mean “no.”]}
@@ -3071,8 +3075,8 @@ \subsubsection{Exercise 52}
 \subsubsection{Exercise 53} $\sim(({\sim p} \wedge q) \vee ({\sim p} \wedge {\sim
 q})) \vee (p \wedge q) \equiv p$
 
-\begin{proof} $\sim(({\sim p} \wedge q) \vee ({\sim p} \wedge {\sim q})) \vee (p
-\wedge q)$
+\begin{proof}
+$\sim(({\sim p} \wedge q) \vee ({\sim p} \wedge {\sim q})) \vee (p \wedge q)$
 
 \begin{tabular}{rcll}
 & $\equiv$ & $\sim({\sim p} \wedge (q \vee {\sim q})) \vee
@@ -3760,14 +3764,20 @@ \subsubsection{Exercise 28}
 
 \subsubsection{Exercise 29}
 $p \to (q \vee r) \equiv (p \wedge {\sim q}) \to r$
-\begin{proof} The tautology is $(p \to (q \vee r)) \bic ((p \wedge {\sim q}) \to
-r)$. $$ \begin{array}{|ccc|c|c|c|c|c|c|} \hline p & q & r & q \vee r & p \wedge
+\begin{proof}
+The tautology is $(p \to (q \vee r)) \bic ((p \wedge {\sim q}) \to r)$. 
+$$ 
+\begin{array}{|ccc|c|c|c|c|c|c|} 
+\hline p & q & r & q \vee r & p \wedge
 {\sim q} & p \to (q \vee r) & p \wedge {\sim q} \to r & (p \to (q \vee r)) \bic
 ((p \wedge {\sim q}) \to r) \\ \hline T & T & T & T & F & T & T & T \\ \hline T
 & T & F & T & F & T & T & T \\ \hline T & F & T & T & T & T & T & T \\ \hline T
 & F & F & F & T & F & F & T \\ \hline F & T & T & T & F & T & T & T \\ \hline F
 & T & F & T & F & T & T & T \\ \hline F & F & T & T & F & T & T & T \\ \hline F
-& F & F & F & F & T & T & T \\ \hline \end{array} $$ \end{proof}
+& F & F & F & F & T & T & T \\ \hline 
+\end{array}
+$$
+\end{proof}
 
 \subsubsection{Exercise 30}
 $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
@@ -4400,7 +4410,9 @@ \subsubsection{Exercise 13}
 
 $\therefore {\sim p}$
 
-\begin{proof} $$ \begin{array}{|cc|c|c|c|}
+\begin{proof}
+$$
+\begin{array}{|cc|c|c|c|}
 \hline
 & & \text{premise} & \text{premise} & \text{conclusion} \\ \hline p & q & p \to q & {\sim q} & {\sim p} \\
 \hline
@@ -4628,7 +4640,8 @@ \subsubsection{Exercise 22}
 
 $\therefore$ Tom is not on team A or Hua is not on team B.
 
-\begin{proof} Let $p$ represent “Tom is on team A” and $q$ represent “Hua is on team B.” Then the argument has the form
+\begin{proof}
+Let $p$ represent “Tom is on team A” and $q$ represent “Hua is on team B.” Then the argument has the form
 
 \begin{center}
 ${\sim p} \to q$ \\
@@ -5111,9 +5124,10 @@ \subsubsection{Exercise 42}
 
 {\bf f.} $\therefore t$
 
-\begin{proof} \begin{tabular}{rrll}
-(1) & & $q \to r$ & by premise (d) \\
-& & ${\sim r}$ & by premise (b) \\
+\begin{proof}
+\begin{tabular}{rrll}
+(1) & & $q \to r$ & by premise (b) \\
+& & ${\sim r}$ & by premise (d) \\
 & $\therefore$ & ${\sim q}$ & by modus tollens \\
 (2) & & ${\sim q}$ & by (1) \\
 & & $p \vee q$ & by premise (a) \\
@@ -5122,7 +5136,7 @@ \subsubsection{Exercise 42}
 & & ${\sim q} \to u \wedge s$ & by premise (e) \\
 & $\therefore$ & $u \wedge s$ & by modus ponens \\
 (4) & & $u \wedge s$ & by (3) \\
-& $\therefore$ & $ $ & by specialization \\
+& $\therefore$ & $s$ & by specialization \\
 (5) & & $p$ & by (2) \\ & & $s$ & by (4) \\
 & $\therefore$ & $p \wedge s$ & by conjunction \\
 (6) & & $p \wedge s$ & by (5) \\
@@ -5201,7 +5215,7 @@ \subsubsection{Exercise 44}
 & $\therefore$ & ${\sim p} \wedge r$ & by conjunction \\
 (7) & & ${\sim p} \wedge r$ & by (6) \\
 & & ${\sim p} \wedge r \to u$ & by premise (f) \\
-& $\therefore$ & $ $ & by modus ponens \\
+& $\therefore$ & $u$ & by modus ponens \\
 (8) & & $u$ & by (7) \\
  & & $w$ & by (2) \\
 & $\therefore$ & $u \wedge w$ & by conjunction \\
@@ -5945,7 +5959,8 @@ \subsubsection{Exercise 33}
 \equiv & {\sim (P \wedge Q)} | {\sim (P \wedge Q)} & \text{\color{cyan}by definition of Sheffer stroke} \\
 \equiv & {\sim ({\sim (P \wedge Q)} \wedge {\sim (P \wedge Q)})} & \text{\color{cyan}by definition of Sheffer stroke} \\
 \equiv & (P \wedge Q) \vee (P \wedge Q) & \text{\color{cyan}by De Morgan laws} \\
-\equiv & P \wedge Q & \text{\color{cyan}by idempotent law} \\ \end{array}
+\equiv & P \wedge Q & \text{\color{cyan}by idempotent law}
+\end{array}
 $$
 \end{proof}
 
@@ -5975,7 +5990,8 @@ \subsubsection{Exercise 34}
 \begin{array}{cll}
 \equiv & {\sim(P \vee P)} & \text{\color{cyan}by definition of Peirce arrow} \\
 \equiv & {\sim P} \wedge {\sim P} & \text{\color{cyan}by De Morgan laws} \\
-\equiv & {\sim P} & \text{\color{cyan}by idempotent law} \\ \end{array}
+\equiv & {\sim P} & \text{\color{cyan}by idempotent law}
+\end{array}
 $$
 \end{proof}
 
@@ -6311,7 +6327,8 @@ \subsubsection{Exercise 23}
 \subsubsection{Exercise 24}
 $-67$
 
-\begin{proof} $|-67|_\base{10} = 67_\base{10} = 64 + 2 + 1 = 01000011_\base{2} \to$ flip the bits $\to 10111100_\base{2} \to$ add 1 $\to 10111101_\base{2}$.
+\begin{proof}
+$|-67|_\base{10} = 67_\base{10} = 64 + 2 + 1 = 01000011_\base{2} \to$ flip the bits $\to 10111100_\base{2} \to$ add 1 $\to 10111101_\base{2}$.
 \end{proof}
 
 \subsubsection{Exercise 25}
@@ -13099,7 +13116,7 @@ \subsubsection{Exercise 30}
 \end{proof}
 
 (b)
-Use the$ $mod$ $notation to rewrite the result of part (a).
+Use the $mod$ notation to rewrite the result of part (a).
 
 \begin{proof}
 For any integer $n$, $n(n+1) = 0 \mod 3$ or $n(n+1) = 2 \mod 3$.
@@ -21279,7 +21296,29 @@ \subsubsection{Exercise 24}
 if one starts from person \#1 and goes repeatedly around the circle successively eliminating every second person, eventually only person \#$(2m + 1)$ will remain.
 
 \begin{proof}
-{\it ???}
+We observe that after eliminating $m$ people (every second person, 
+starting from \#1), we will have removed the people numbered 
+$\#2, \#4, \ldots,\#(2m)$ leaving person \#$(2m+1)$ next in line.
+
+We also know that:
+\begin{align*}
+    2^n+m &< 2^{n+1} \\
+    m &< 2^n \\
+    m+1 &\le 2^n \\
+    2m+1 &\le 2^n+m
+\end{align*}
+
+In other words, since $2m+1 \in [1,r]$, the elimination never loops 
+through the circle more than once; it terminates within the first 
+traversal. This means we don’t need to worry about adjusting for shifts 
+caused by previously eliminated people in a second pass.
+
+After eliminating $m$ people, there are $2^n$ people remaining. 
+By part (b), when $2^n$ people remain in the circle and we start from 
+the current person — in this case, person \#$(2m + 1)$ — the last 
+remaining person will be that same person.
+
+Therefore, the last remaining person is \#$(2m + 1)$, as required.
 \end{proof}
 
 \subsubsection{Exercise 25}