279 lines
9.3 KiB
TeX
279 lines
9.3 KiB
TeX
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\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 1}
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\begin{document}
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\ntnuTitle{}
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\break{}
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\begin{excs}
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\exc{}
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\begin{truthtable}
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{c|c|c}
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{$p$ & $q$ & $p \Rightarrow q$}
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\T & \T & \T \\
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\erow{}
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\T & \F & \F \\
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\F & \T & \T \\
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\F & \F & \T \\
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\end{truthtable}
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Looking at the truthtable, we can see that $p \Rightarrow q$ only is false when $p$ is true and $q$ is false.
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\begin{subexcs}
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\subexc{}
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\begin{align*}
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p \wedge q &\equiv T \wedge F \\
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&\equiv F
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\end{align*}
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\subexc{}
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\begin{align*}
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\neg p \vee q &\equiv \neg T \vee F \\
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&\equiv F \vee F \\
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&\equiv F
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\end{align*}
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\subexc{}
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\begin{align*}
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q \Rightarrow p &\equiv F \Rightarrow T \\
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&\equiv \neg F \vee T \\
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&\equiv T \vee T \\
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&\equiv T
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\end{align*}
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\subexc{}
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\begin{align*}
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\neg q \Rightarrow \neg p &\equiv \neg F \Rightarrow \neg T \\
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&\equiv T \Rightarrow F \\
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&\equiv \neg T \vee F \\
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&\equiv F \vee F \\
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&\equiv F
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\end{align*}
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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If triangle ABC is equilateral then triangle ABC is isosceles.
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\subexc{}
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If triangle ABC is not isosceles then triangle ABC is not equilateral.
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\subexc{}
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Triangle ABC is equilateral if and only if triangle ABC is equiangular.
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\subexc{}
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Triangle ABC is isosceles and triangle ABC is not equilateral.
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\subexc{}
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If triangle ABC is equiangular then triangle ABC is isosceles.
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|c|c|c}
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{$p$ & $q$ & $\neg p$ & $\neg q$ & $p \wedge \neg q$ & $\neg (p \wedge \neg q)$ & $\neg (p \wedge \neg q) \Rightarrow p$}
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\T & \T & \F & \F & \F & \T & \F \\
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\T & \F & \F & \T & \T & \F & \T \\
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\F & \T & \T & \F & \F & \T & \T \\
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\F & \F & \T & \T & \F & \T & \T
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\end{truthtable}
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|c}
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{$p$ & $q$ & $r$ & $q \Rightarrow r$ & $p \Rightarrow (q \Rightarrow r)$}
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\T & \T & \T & \T & \T \\
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\T & \T & \F & \F & \F \\
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\T & \F & \T & \T & \T \\
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\T & \F & \F & \T & \T \\
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\F & \T & \T & \T & \T \\
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\F & \T & \F & \F & \T \\
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\F & \F & \T & \T & \T \\
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\F & \F & \F & \T & \T
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\end{truthtable}
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|c|e}
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{$p$ & $q$ & $\neg p$ & $\neg q$ & $\neg p \vee \neg q$ & $q \Leftrightarrow (\neg p \vee \neg q)$}
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\T & \T & \F & \F & \F & \F \\
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\T & \F & \F & \T & \T & \F \\
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\F & \T & \T & \F & \T & \T \\
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\F & \F & \T & \T & \T & \F \\
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\end{truthtable}
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$q \Leftrightarrow (\neg p \vee \neg q)$ is not a tautology.
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|c|c|c|e}
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{$p$ & $q$ & $r$ & $p \Rightarrow q$ & $q \Rightarrow r$ & $p \Rightarrow r$ & $(p \Rightarrow q) \wedge (q \Rightarrow r)$ & $\left[ (p \Rightarrow q) \wedge (q \Rightarrow r) \right] \Rightarrow (p \Rightarrow r)$}
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\T & \T & \T & \T & \T & \T & \T & \T \\
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\T & \T & \F & \T & \F & \F & \F & \T \\
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\T & \F & \T & \F & \T & \T & \F & \T \\
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\T & \F & \F & \F & \T & \F & \F & \T \\
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\F & \T & \T & \T & \T & \T & \T & \T \\
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\F & \T & \F & \T & \F & \T & \F & \T \\
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\F & \F & \T & \T & \T & \T & \T & \T \\
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\F & \F & \F & \T & \T & \T & \T & \T
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\end{truthtable}
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$\left[ (p \Rightarrow q) \wedge (q \Rightarrow r) \right] \Rightarrow (p \Rightarrow r)$ is a tautology.
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\end{subexcs}
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\exc{}
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I start by simplifying the expression, inserting $q$ as $T$
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\begin{align*}
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\left(q \Rightarrow \left[ (\neg p \vee r) \wedge \neg s \right]\right) \wedge \left[\neg s \Rightarrow (\neg r \wedge q)\right] &\equiv
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\left(T \Rightarrow \left[ (\neg p \vee r) \wedge \neg s \right]\right) \wedge \left[\neg s \Rightarrow (\neg r \wedge T)\right] \\
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&\equiv \left(\neg T \vee \left[ (\neg p \vee r) \wedge \neg s \right]\right) \wedge \left[\neg s \Rightarrow \neg r \right] \\
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&\equiv \left(F \vee \left[ (\neg p \vee r) \wedge \neg s \right]\right) \wedge \left[\neg s \Rightarrow \neg r \right] \\
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&\equiv \left[ (\neg p \vee r) \wedge \neg s \right] \wedge \left[\neg s \Rightarrow \neg r \right] \\
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\end{align*}
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\begin{truthtable}
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{c|c|c|c|c|c|c|c|c|c}
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{$p$ & $r$ & $s$ & $\neg p$ & $\neg r$ & $\neg s$ & $\neg p \vee r$ & $(\neg p \vee r) \wedge \neg s$ & $\neg s \Rightarrow \neg r$ & $\left[ (\neg p \vee r) \wedge \neg s \right] \wedge \left[\neg s \Rightarrow \neg r \right]$}
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\T & \T & \T & \F & \F & \F & \T & \F & \T & \F \\
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\T & \T & \F & \F & \F & \T & \T & \T & \F & \F \\
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\T & \F & \T & \F & \T & \F & \F & \F & \T & \F \\
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\T & \F & \F & \F & \T & \T & \F & \F & \T & \F \\
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\F & \T & \T & \T & \F & \F & \T & \F & \T & \F \\
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\F & \T & \F & \T & \F & \T & \T & \T & \F & \F \\
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\F & \F & \T & \T & \T & \F & \T & \F & \T & \F \\
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\erow{}
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\F & \F & \F & \T & \T & \T & \T & \T & \T & \T
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\end{truthtable}
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The statement is only true when $p$, $r$ and $s$ are false.
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|e|c|c|e}
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{$p$ & $q$ & $r$ & $q \wedge r$ & $p \Rightarrow (q \wedge r)$ & $p \Rightarrow q$ & $p \Rightarrow r$ & $(p \Rightarrow q) \wedge (p \Rightarrow r)$}
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\T & \T & \T & \T & \T & \T & \T & \T \\
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\T & \T & \F & \F & \F & \T & \F & \F \\
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\T & \F & \T & \F & \F & \F & \T & \F \\
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\T & \F & \F & \F & \F & \F & \F & \F \\
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\F & \T & \T & \F & \T & \T & \T & \T \\
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\F & \T & \F & \F & \T & \T & \T & \T \\
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\F & \F & \T & \F & \T & \T & \T & \T \\
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\F & \F & \F & \F & \T & \T & \T & \T
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\end{truthtable}
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\subexc{}
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\begin{truthtable}
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{c|c|c|c|e|c|c|e}
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{$p$ & $q$ & $r$ & $q \vee r$ & $p \Rightarrow (q \vee r)$ & $\neg r$ & $p \Rightarrow q$ & $\neg r \Rightarrow (p \Rightarrow q)$}
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\T & \T & \T & \T & \T & \F & \T & \T \\
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\T & \T & \F & \T & \T & \T & \T & \T \\
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\T & \F & \T & \T & \T & \F & \F & \T \\
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\T & \F & \F & \F & \F & \T & \F & \F \\
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\F & \T & \T & \T & \T & \F & \T & \T \\
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\F & \T & \F & \T & \T & \T & \T & \T \\
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\F & \F & \T & \T & \T & \F & \T & \T \\
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\F & \F & \F & \F & \T & \T & \T & \T
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\end{truthtable}
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{align*}
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\neg((p \wedge q) \Rightarrow r) &\equiv \neg(\neg(p \wedge q) \vee r) \\
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&\equiv \neg\neg(p \wedge q) \wedge \neg r \\
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&\equiv (p \wedge q) \wedge \neg r \\
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&\equiv p \wedge q \wedge \neg r \\
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\end{align*}
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\subexc{}
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\begin{align*}
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\neg(p \Rightarrow (\neg q \wedge r)) &\equiv \neg(\neg p \vee (\neg q \wedge r)) \\
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&\equiv \neg\neg p \wedge \neg(\neg q \wedge r) \\
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&\equiv p \wedge (\neg\neg q \vee \neg r) \\
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&\equiv p \wedge (q \vee \neg r)
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\end{align*}
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\end{subexcs}
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\exc{}
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\begin{truthtable}
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{c|c|c|c|c|e|e}
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{$\alpha$ & $\beta$ & $\gamma$ & $\alpha \vee \beta$ & $\beta \vee \gamma$ & $(\alpha \vee \beta) \vee \gamma$ & $\alpha \vee (\beta \vee \gamma)$}
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\T & \T & \T & \T & \T & \T & \T \\
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\T & \T & \F & \T & \T & \T & \T \\
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\T & \F & \T & \T & \T & \T & \T \\
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\T & \F & \F & \T & \F & \T & \T \\
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\F & \T & \T & \T & \T & \T & \T \\
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\F & \T & \F & \T & \T & \T & \T \\
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\F & \F & \T & \F & \T & \T & \T \\
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\F & \F & \F & \F & \F & \F & \F
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\end{truthtable}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{truthtable}
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{c|c|c|e}
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{$p$ & $q$ & $p \vee q$ & $p \Rightarrow (p \vee q)$}
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\T & \T & \T & \T \\
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\T & \F & \T & \T \\
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\F & \T & \T & \T \\
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\F & \F & \F & \T \\
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\end{truthtable}
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$p \Rightarrow (p \vee q)$ is a tautology
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\subexc{} Because $\neg(p \Rightarrow (p \vee q))$ is the negation of $p \Rightarrow (p \vee q)$, which we have already evaluated to be a tautology, this has to be a contradiction and thus unsatisfiable.
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\subexc{}
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\begin{truthtable}
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{c|c|c|e}
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{$p$ & $q$ & $p \Rightarrow q$ & $p \Rightarrow (p \Rightarrow q)$}
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\T & \T & \T & \T \\
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\T & \F & \F & \F \\
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\F & \T & \T & \T \\
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\F & \F & \T & \T \\
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\end{truthtable}
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$p \Rightarrow (p \Rightarrow q)$ is satisfiable.
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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$\neg p \Rightarrow (q \Leftrightarrow r)$
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\subexc{}
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$r \Rightarrow \neg p$
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\subexc{}
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$\neg r \wedge (p \wedge q)$
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\subexc{}
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$p \Rightarrow (r \wedge q)$
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\subexc{}
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$\neg q \wedge r$
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\end{subexcs}
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\end{excs}
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\end{document}
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