Skip to content

add proof for 5.4.24.c, formatting and minor fixes #30

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 2 commits into from
Aug 6, 2025
Merged

add proof for 5.4.24.c, formatting and minor fixes #30

Changes from all commits
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
b1d88b0
add proof for 5.4.24.c, formatting and minor fixes
tenick Aug 6, 2025
fe76b7a
update align
tenick Aug 6, 2025
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
81 changes: 60 additions & 21 deletions src/Epp.tex
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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}

Expand Down Expand Up @@ -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}
Expand Down Expand Up @@ -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.
Expand Down Expand Up @@ -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.”]}
Expand Down Expand Up @@ -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
Expand Down Expand Up @@ -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}
$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
Expand Down Expand Up @@ -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
Expand Down Expand Up @@ -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}
${\sim p} \to q$ \\
Expand Down Expand Up @@ -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) \\
Expand All @@ -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) \\
Expand Down Expand Up @@ -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 \\
Expand Down Expand Up @@ -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}

Expand Down Expand Up @@ -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}

Expand Down Expand Up @@ -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}
Expand Down Expand Up @@ -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 $n(n+1) = 2 \mod 3$.
Expand Down Expand Up @@ -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}
Expand Down
Loading