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*.prv
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auto
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_region_.tex
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*.pdf
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*.aux
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*.fdb_latexmk
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*.fls
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*.log
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*.out
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*.synctex.gz
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.projectile
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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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@ -0,0 +1,119 @@
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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 2}
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||||
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\begin{document}
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||||
\ntnuTitle{}
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\break{}
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||||
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||||
\begin{excs}
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\exc{}
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||||
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||||
\begin{truthtable}
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{e|c|c|e}
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{$p$ & $q$ & $p \wedge q$ & $p \vee (p \wedge q)$}
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||||
\T & \T & \T & \T \\
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\T & \F & \F & \T \\
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\F & \T & \F & \F \\
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\F & \F & \F & \F \\
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\end{truthtable}
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||||
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\exc{}
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(DL1)
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\begin{truthtable}
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||||
{c|c|c|c|e|c|c|e}
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{$\alpha$ & $\beta$ & $\gamma$ & $(\beta \wedge \gamma)$ & $\alpha \vee (\beta \wedge \gamma)$ & $\alpha \vee \beta$ & $\alpha \vee \gamma$ & $(\alpha \vee \gamma) \wedge (\alpha \vee \gamma)$}
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||||
\T & \T & \T & \T & \T & \T & \T & \T \\
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||||
\T & \T & \F & \F & \T & \T & \T & \T \\
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||||
\T & \F & \T & \F & \T & \T & \T & \T \\
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||||
\T & \F & \F & \F & \T & \T & \T & \T \\
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||||
\F & \T & \T & \T & \T & \T & \T & \T \\
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||||
\F & \T & \F & \F & \F & \T & \F & \F \\
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||||
\F & \F & \T & \F & \F & \F & \T & \F \\
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||||
\F & \F & \F & \F & \F & \F & \F & \F \\
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||||
\end{truthtable}
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||||
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(DL2)
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||||
\begin{truthtable}
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||||
{c|c|c|c|e|c|c|e}
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||||
{$\alpha$ & $\beta$ & $\gamma$ & $(\beta \vee \gamma)$ & $\alpha \wedge (\beta \vee \gamma)$ & $\alpha \wedge \beta$ & $\alpha \wedge \gamma$ & $(\alpha \wedge \gamma) \vee (\alpha \wedge \gamma)$}
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||||
\T & \T & \T & \T & \T & \T & \T & \T \\
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||||
\T & \T & \F & \T & \T & \T & \F & \T \\
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||||
\T & \F & \T & \T & \T & \F & \T & \T \\
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||||
\T & \F & \F & \F & \F & \F & \F & \F \\
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||||
\F & \T & \T & \T & \F & \F & \F & \F \\
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||||
\F & \T & \F & \T & \F & \F & \F & \F \\
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||||
\F & \F & \T & \T & \F & \F & \F & \F \\
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||||
\F & \F & \F & \F & \F & \F & \F & \F \\
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||||
\end{truthtable}
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||||
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||||
\exc{}
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||||
\begin{align*}
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||||
p \Rightarrow (q \vee r) &\equiv (p \wedge \neg q) \Rightarrow r \\
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||||
&\equiv \neg (p \wedge \neg q) \vee r \\
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||||
&\equiv (\neg p \vee \neg\neg q) \vee r \\
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||||
&\equiv (\neg p \vee q) \vee r \\
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||||
&\equiv \neg p \vee (q \vee r) \\
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||||
&\equiv p \Rightarrow (q \vee r)
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||||
\end{align*}
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||||
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||||
\exc{}
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||||
\begin{align*}
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||||
[(q \wedge p) \vee q] \wedge \neg(\neg q \vee p) &\equiv q \wedge \neg p \\
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||||
q \wedge \neg p &\equiv [(q \wedge p) \vee q] \wedge \neg(\neg q \vee p) \\
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||||
&\equiv [q] \wedge (\neg\neg q \wedge \neg p) \\
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||||
&\equiv q \wedge (q \wedge \neg p) \\
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||||
&\equiv (q \wedge q) \wedge \neg p \\
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||||
&\equiv q \wedge \neg p
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||||
\end{align*}
|
||||
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||||
\exc{}
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||||
\begin{subexcs}
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||||
\subexc{}
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||||
$\forall S(x)[H(x)]$
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||||
\subexc{}
|
||||
$\exists S(x)[\neg H(x)]$
|
||||
\subexc{}
|
||||
$\forall S(x)[\neg H(x)]$
|
||||
\subexc{}
|
||||
$\forall \neg H(x) \exists S(x)$
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
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||||
\exc{}
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||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
The formula is true because of the case where $x < z < y$ which would mean that $x < y \wedge z < y \wedge x < z \wedge \neg (z < x)$
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||||
\subexc{}
|
||||
The formula is false because $p(z, y)$ and $\neg p(z, x)$ cannot be fulfilled at the same time. $z \geq 0 \wedge \neg (z \geq 0) \equiv F$
|
||||
|
||||
\setsubexc{4}
|
||||
\subexc{}
|
||||
\fbox{see comment in ovsys}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
\neg (\forall x [p(x) \wedge q(x)]) &\equiv \exists x [\neg(p(x) \wedge q(x))] \\
|
||||
&\equiv \exists x [\neg p(x) \vee \neg q(x)]
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
\neg (\exists x \forall y [p(y) \vee \neg q(x,y)]) &\equiv \forall x \neg(\forall y [p(y) \vee \neg q(x,y)]) \\
|
||||
&\equiv \forall x \exists y \neg[p(y) \vee \neg q(x,y)] \\
|
||||
&\equiv \forall x \exists y [\neg p(y) \wedge \neg\neg q(x,y)] \\
|
||||
&\equiv \forall x \exists y [\neg p(y) \wedge q(x,y)]
|
||||
\end{align*}
|
||||
|
||||
\end{excs}
|
||||
|
||||
|
||||
\end{document}
|
|
@ -0,0 +1,282 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 3}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
\neg ((\neg p \wedge q) \vee (\neg p \wedge \neg q)) &\vee (p \wedge q) && \\
|
||||
\neg (\neg p \wedge (q \vee \neg q)) &\vee (p \wedge q) && \text{Distributive law} \\
|
||||
\neg (\neg p \wedge T) &\vee (p \wedge q) && \text{Complement law} \\
|
||||
\neg (\neg p) &\vee (p \wedge q) && \text{Identity law} \\
|
||||
p &\vee (p \wedge q) && \text{Double negation law} \\
|
||||
p & && \text{Absortion law} \\
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
((p \wedge q) \vee (p \wedge \neg r) \vee \neg(\neg p \vee q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \\
|
||||
((p \wedge q) \vee (p \wedge \neg r) \vee (\neg\neg p \wedge \neg q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{De Morgans's law} \\
|
||||
((p \wedge q) \vee (p \wedge \neg r) \vee ( p \wedge \neg q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Double negation law} \\
|
||||
((p \wedge (q \vee \neg q)) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Distributive law} \\
|
||||
((p \wedge T) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Complement law} \\
|
||||
((p) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Identity law} \\
|
||||
p &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Absortion law} \\
|
||||
p &\vee ((\neg q \wedge r) \vee (\neg q \wedge s) \vee (\neg q \wedge \neg r)) && \text{Distributive law} \\
|
||||
p &\vee (\neg q \wedge r) \vee (\neg q \wedge s) \vee (\neg q \wedge \neg r) && \text{Associative law}
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
|
||||
\renewcommand{\theenumii}{\roman{enumii})}
|
||||
\renewcommand{\theenumiii}{\alph{enumiii})}
|
||||
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
\begin{ssubexcs}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
\{\{2,3,5\} \cup \{6,4\}\} \cap \{4,6,8\} \\
|
||||
\{\{2,4,6\}\} \cap \{4,6,8\} \\
|
||||
\emptyset
|
||||
\end{gather*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
P(\{7,8,9\}) &- P(\{7,9\}) \\
|
||||
\{\{7,8,9\}, \{7,8\}, \{8,9\}, \{7,9\}, \{7\}, \{8\}, \{9\}, \emptyset\} &- \{\{7,9\}, \{7\}, \{9\}, \emptyset\} \\
|
||||
\{\{7,8,9\}, \{7,8\}, \{8,9\}, \{8\}\} & \\
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
P(\emptyset) \\
|
||||
\{\emptyset\}
|
||||
\end{gather*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
\{1, 3, 5\} \times \{0\} \\
|
||||
\{ \langle 1,0 \rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle \}
|
||||
\end{gather*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
\{2,4,6\} \times \emptyset \\
|
||||
\emptyset
|
||||
\end{gather*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
P(\{0\}) \times P(\{1\}) \\
|
||||
\{\emptyset, \{0\}\} \times \{\emptyset, \{1\}\} \\
|
||||
\{\langle\emptyset,\emptyset\rangle, \langle\emptyset,\{1\}\rangle, \langle\{0\},\emptyset\rangle, \langle\{0\},\{1\}\rangle\}
|
||||
\end{gather*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{gather*}
|
||||
P(P(\{2\})) \\
|
||||
P(\{\emptyset,\{2\}\}) \\
|
||||
\{ \{\{\emptyset\}, \{2\}\}, \{\{\emptyset\}\}, \{\{2\}\}, \emptyset \}
|
||||
\end{gather*}
|
||||
|
||||
\end{ssubexcs}
|
||||
|
||||
\subexc{}
|
||||
Because the elements in a power set can be represented as a binary tree where every leaf node is a set that has the cardinality of $1$, and that $\{\{x\} : x \in A\}$ would make up all the leaf nodes, we can reason that
|
||||
\[ |P(A) - \{\{x\} : x \in A\}| = \frac{n}{2} \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\renewcommand{\theenumii}{\alph{enumii})}
|
||||
\renewcommand{\theenumiii}{\roman{enumiii})}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
$\emptyset = \{\emptyset\}$ is {\color{red}False} because $|\emptyset| \neq |\{\emptyset\}|$
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = \{0\}$ is {\color{red}False} because $|\emptyset| \neq |\{0\}|$
|
||||
|
||||
\subexc{}
|
||||
$|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
|
||||
|
||||
\subexc{}
|
||||
$P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = \{\{\emptyset\}\}$ has $1$ element
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = \{\}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = \{x \in \mathbb{N} : x \leq 0 and x > 0\}$ is {\color{red}False} because $x \leq 0 \wedge x > 0 \equiv \F$, which means there are no such elements, and thus the set is empty
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
&A \cap (A \cup B) \\
|
||||
&\{x : x \in A \wedge x \in (A \cup B)\} \\
|
||||
&\{x : x \in A \wedge (x \in A \vee x \in B)\} \\
|
||||
&\{x : x \in A\} \\
|
||||
&A
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
&A-(B \cap C) \\
|
||||
&\{x : x \in A \wedge x \notin (B \cap C)\} \\
|
||||
&\{x : x \in A \wedge (x \notin B \wedge x \notin C)\} \\
|
||||
&\{x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)\} \\
|
||||
&\{x : x \in (A - B) \vee x \in (A - C)\} \\
|
||||
&\{x : x \in (A - B) \cup (A - C)\} \\
|
||||
&(A-B) \cup (A-C)
|
||||
\end{align*}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
|
||||
\renewcommand{\theenumii}{\roman{enumii})}
|
||||
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
\begin{align*}
|
||||
&(A \cup B) \setminus (A \cap B) \\
|
||||
&\{x: x \in (A \cup B) \setminus (A \cap B)\} \\
|
||||
&\{x: x \in (A \cup B) \wedge x \notin (A \cap B)\} \\
|
||||
&\{x: (x \in A \vee x \in B) \wedge (x \notin A \vee x \notin B)\} \\
|
||||
&\{x: x \in A \wedge (x \notin A \vee x \notin B) \vee x \in B \wedge (x \notin A \vee x \notin B) \} \\
|
||||
&\{x: ((x \in A \wedge x \notin A) \vee (x \in A \wedge x \notin B)) \vee ((x \in B \wedge x \notin A) \vee (x \in B \wedge x \notin B)) \} \\
|
||||
&\{x: (F \vee (x \in A \wedge x \notin B)) \vee ((x \in B \wedge x \notin A) \vee F) \} \\
|
||||
&\{x: (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin A) \} \\
|
||||
&\{x: x \in (A - B) \vee x \in (B - A) \} \\
|
||||
&\{x: x \in (A - B) \cup (B - A) \} \\
|
||||
&(A - B) \cup (B - A) \\
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
For this exercise, I counted the elements which was in either set but not both
|
||||
\[ A \Delta B = \{2, 4, 6, 7, 8\} \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\renewcommand{\theenumii}{\alph{enumii})}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
X &= \{\{1,2,3\}, \{2,3\}, \{ef\}\} \cup \{\{e\}\} \\
|
||||
&= \{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\} \\
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
P(x) = \{ \\
|
||||
&\{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\}, \\
|
||||
&\{\{1,2,3\}, \{2,3\}, \{ef\}\}, \\
|
||||
&\{\{1,2,3\}, \{2,3\}, \{e\}\}, \\
|
||||
&\{\{1,2,3\}, \{ef\}, \{e\}\}, \\
|
||||
&\{\{2,3\}, \{ef\}, \{e\}\}, \\
|
||||
&\{\{1,2,3\}, \{2,3\}\}, \\
|
||||
&\{\{1,2,3\}, \{e\}\}, \\
|
||||
&\{\{1,2,3\}, \{ef\}\}, \\
|
||||
&\{\{2,3\}, \{ef\}\}, \\
|
||||
&\{\{2,3\}, \{e\}\}, \\
|
||||
&\{\{ef\}, \{e\}\}, \\
|
||||
&\{\{e\}\}, \\
|
||||
&\{\{ef\}\}, \\
|
||||
&\{\{2,3\}\}, \\
|
||||
&\{\{1,2,3\}\} \\
|
||||
\} \\
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
P(X \cap Y) &= P(\{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\} \cap \{\{1,2,3,e,f\}\}) \\
|
||||
&= P(\emptyset) \\
|
||||
&= \{\emptyset\}
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
Here, the exercise says ``[\ldots] four sets $A_1$, $A_2$, $A_3$''. I'm not sure if I'm supposed to do three or four, but I'll assume three sets $A_1$, $A_2$, $A_3$ was the intention.
|
||||
|
||||
\begin{align*}
|
||||
A_1 \cap A_2 \cap A_3 \\
|
||||
A_1 \cap A_2 \cap \overline{A_3} \\
|
||||
A_1 \cap \overline{A_2} \cap A_3 \\
|
||||
A_1 \cap \overline{A_2} \cap \overline{A_3} \\
|
||||
\overline{A_1} \cap A_2 \cap A_3 \\
|
||||
\overline{A_1} \cap A_2 \cap \overline{A_3} \\
|
||||
\overline{A_1} \cap \overline{A_2} \cap A_3 \\
|
||||
\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{gather*}
|
||||
A \overline{( B \overline{C} )} \overline{( (A \overline{B}) \overline{C})} \\
|
||||
A ( \overline{B} + \overline{\overline{C}} ) \overline{(A \overline{B}\ \overline{C})} \\
|
||||
A ( \overline{B} + C ) \overline{(A \overline{B}\ \overline{C})} \\
|
||||
A ( \overline{B} + C ) \overline{(A \overline{B}\ \overline{C})} \\
|
||||
A ( \overline{B} + C ) (\overline{A} + \overline{\overline{B}} + \overline{\overline{C}}) \\
|
||||
A ( \overline{B} + C ) (\overline{A} + B + C) \\
|
||||
( A\overline{B} + AC ) (\overline{A} + B + C) \\
|
||||
A\overline{B}(\overline{A} + B + C) + AC(\overline{A} + B + C)\\
|
||||
(A\overline{B}\ \overline{A} + A\overline{B}B + A\overline{B}C) + (AC\overline{A} + ACB + ACC)\\
|
||||
(0 + 0 + A\overline{B}C) + (0 + ACB + AC)\\
|
||||
A\overline{B}C + ACB + AC\\
|
||||
ACB + AC\\
|
||||
AC
|
||||
\end{gather*}
|
||||
|
||||
\exc{}
|
||||
LHS
|
||||
\begin{gather*}
|
||||
((A+B)+(A+C)) \overline{((A+B)(A+C))} \overline{A} \\
|
||||
(A+B+A+C) \overline{(A+B)(A+C)} \overline{A} \\
|
||||
(A+B+C) (\overline{(A+B)} + \overline{(A+C)}) \overline{A} \\
|
||||
(A+B+C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \overline{A} \\
|
||||
(\overline{A}A + \overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
|
||||
(0 + \overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
|
||||
(\overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
|
||||
\overline{A}B \overline{A}\ \overline{B} +
|
||||
\overline{A}B \overline{A}\ \overline{C} +
|
||||
\overline{A}C \overline{A}\ \overline{B} +
|
||||
\overline{A}C \overline{A}\ \overline{C} \\
|
||||
0 +
|
||||
\overline{A}B \overline{A}\ \overline{C} +
|
||||
\overline{A}C \overline{A}\ \overline{B} +
|
||||
0 \\
|
||||
\overline{A}B \overline{C} + \overline{A}C \overline{B} \\
|
||||
\end{gather*}
|
||||
|
||||
RHS
|
||||
\begin{gather*}
|
||||
(B+C) \overline{(BC)} \overline{A} \\
|
||||
(B+C) (\overline{B} + \overline{C}) \overline{A} \\
|
||||
(\overline{A}B + \overline{A}C) (\overline{B} + \overline{C}) \\
|
||||
\overline{A}B\overline{B} + \overline{A}B\overline{C} + \overline{A}C\overline{B} + \overline{A}C\overline{C} \\
|
||||
0 + \overline{A}B\overline{C} + \overline{A}C\overline{B} + 0 \\
|
||||
\overline{A}B\overline{C} + \overline{A}C\overline{B} \\
|
||||
\end{gather*}
|
||||
|
||||
$LHS = RHS$
|
||||
|
||||
\end{excs}
|
||||
|
||||
\end{document}
|
|
@ -0,0 +1,185 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 2}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
\exc{}
|
||||
|
||||
\exc{}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
\begin{ssubexcs}
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
{{2,3,5} \cup {6,4}} &\cap {4,6,8} \\
|
||||
{{2,4,6}} &\cap {4,6,8} \\
|
||||
\emptyset
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
P({7,8,9}) &- P({7,9}) \\
|
||||
{{7,8,9}, {7,8}, {8,9}, {7,9}, {7}, {8}, {9}, \emptyset} &- {{7,9}, {7}, {9}, \emptyset} \\
|
||||
{{7,8,9}, {7,8}, {8,9}, {8}}
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
P(\emptyset) \\
|
||||
{\emptyset}
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
{1, 3, 5} \times {0} \\
|
||||
{ \langle 1,0 \rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle }
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
{2,4,6} \times \emptyset \\
|
||||
\emptyset
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
P({0}) &\times P({1}) \\
|
||||
{\emptyset, {0}} &\times {\emptyset, {1}} \\
|
||||
{\langle\emptyset,\emptyset\rangle, \langle\emptyset,{1}\rangle, \langle{0},\emptyset\rangle, \langle{0},{1}\rangle}
|
||||
\end{align*}
|
||||
|
||||
\ssubexc{}
|
||||
\begin{align*}
|
||||
P(P({2})) \\
|
||||
P({\emptyset,{2}}) \\
|
||||
{ {{\emptyset}, {2}}, {{\emptyset}}, {{2}}, \emptyset }
|
||||
\end{align*}
|
||||
|
||||
\end{ssubexcs}
|
||||
|
||||
\subexc{}
|
||||
Because the elements in a power set can be represented as a binary tree where every leaf node is a set that has the cardinality of $1$, and that ${{x} : x \in A}$ would make up all the leaf nodes, we can reason that
|
||||
\[ |P(A) - {{x} : x \in A}| = \frac{n}{2} \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
$\emptyset = {\emptyset}$ is {\color{red}False} because $|\emptyset| \neq |{\emptyset}|$
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = {0}$ is {\color{red}False} because $|\emptyset| \neq |{0}|$
|
||||
|
||||
\subexc{}
|
||||
$|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
|
||||
|
||||
\subexc{}
|
||||
$P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = {{\emptyset}}$ has $1$ element
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = {}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
|
||||
|
||||
\subexc{}
|
||||
$\emptyset = {x \in \mathbb{N} : x \leq 0 and x > 0}$ is {\color{red}False} because $x \leq 0 \wedge x > 0 \equiv \F$, which means there are no such elements, and thus the set is empty
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
A \cap (\A \cup B) \\
|
||||
{x : x \in A \wedge x \in (A \cup B)} \\
|
||||
{x : x \in A \wedge (x \in A \or x \in B)} \\
|
||||
{x : x \in A} \\
|
||||
A
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
A-(B \cap C) \\
|
||||
{x : x \in A \wedge x \notin (B \cap C)} \\
|
||||
{x : x \in A \wedge (x \notin B \wedge x \notin C)} \\
|
||||
{x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)} \\
|
||||
{x : x \in (A - B) \vee x \in (A - C)} \\
|
||||
{x : x \in (A - B) \cup (A - C)} \\
|
||||
(A-B) \cup (A-C)
|
||||
\end{align*}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
X &= {{1,2,3}, {2,3}, {ef}} \cup {{e}} \\
|
||||
&= {{1,2,3}, {2,3}, {ef}, {e}} \\
|
||||
\\
|
||||
P(x) &= {
|
||||
{{1,2,3}, {2,3}, {ef}, {e}},
|
||||
{{1,2,3}, {2,3}, {ef}},
|
||||
{{1,2,3}, {2,3}, {e}},
|
||||
{{1,2,3}, {ef}, {e}},
|
||||
{{2,3}, {ef}, {e}},
|
||||
{{1,2,3}, {2,3}}
|
||||
{{1,2,3}, {e}}
|
||||
{{1,2,3}, {ef}}
|
||||
{{2,3}, {ef}}
|
||||
{{2,3}, {e}}
|
||||
{{ef}, {e}}
|
||||
{{e}}
|
||||
{{ef}},
|
||||
{{2,3}},
|
||||
{{1,2,3}}
|
||||
} \\
|
||||
\\
|
||||
P(X \cap Y) &= P({{1,2,3}, {2,3}, {ef}, {e}} \cap {{1,2,3,e,f}}) \\
|
||||
&= P(\emptyset) \\
|
||||
&= {\emptyset}
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
Here, the exercise says ``[\ldots] four sets $A_1$, $A_2$, $A_3$''. I'm not sure if I'm supposed to do three or four, but I'll assume three.
|
||||
|
||||
\begin{align*}
|
||||
A_1 \cap A_2 \cap A_3 \\
|
||||
A_1 \cap A_2 \cap \overline{A_3} \\
|
||||
A_1 \cap \overline{A_2} \cap A_3 \\
|
||||
A_1 \cap \overline{A_2} \cap \overline{A_3} \\
|
||||
\overline{A_1} \cap A_2 \cap A_3 \\
|
||||
\overline{A_1} \cap A_2 \cap \overline{A_3} \\
|
||||
\overline{A_1} \cap \overline{A_2} \cap A_3 \\
|
||||
\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{align*}
|
||||
A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
|
||||
A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
|
||||
\end{align*}
|
||||
|
||||
\end{excs}
|
||||
|
||||
|
||||
\end{document}
|
|
@ -0,0 +1,199 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 4}
|
||||
|
||||
\usepackage{amsthm}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
\overline{xy} + \overline{x}\ \overline{y} \\
|
||||
\overline{1 \cdot 0} + (\overline{1}\cdot \overline{0}) \\
|
||||
\overline{0} + (0 \cdot 1) \\
|
||||
1 + 0 \\
|
||||
1
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
w + \overline{x}y \\
|
||||
1 + (\overline{1} \cdot 0) \\
|
||||
1 + 0 \\
|
||||
1
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
wx + \overline{y} + yz \\
|
||||
(1 \cdot 1) + \overline{0} + (0 \cdot 0) \\
|
||||
1 + 1 + 0 \\
|
||||
1
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
(wx + y\overline{z}) + w\overline{y} + \overline{(w + y)(\overline{x} + y)} \\
|
||||
((1 \cdot 1) + (0 \cdot \overline{0})) + (1 \cdot \overline{0}) + \overline{(1 + 0)(\overline{1} + 0)} \\
|
||||
(1 + 0) + (1 \cdot 1) + \overline{(1)(0 + 0)} \\
|
||||
1 + 1 + \overline{(1)(0)} \\
|
||||
1 + 1 + \overline{0} \\
|
||||
1 + 1 + 1 \\
|
||||
1
|
||||
\end{gather*}
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
xy + (x + y)\overline{z} + y \\
|
||||
(xy + y) + \overline{z}x + \overline{z}y \\
|
||||
y + \overline{z}x + \overline{z}y \\
|
||||
\overline{z}x + (y + \overline{z}y) \\
|
||||
\overline{z}x + y
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
x + y + \overline{(\overline{x} + y + z)} \\
|
||||
x + y + \overline{\overline{x}}\ \overline{y}\ \overline{z} \\
|
||||
x + y + x\overline{y}\ \overline{z} \\
|
||||
(x + x\overline{y}\ \overline{z}) + y \\
|
||||
x + y
|
||||
\end{gather*}
|
||||
|
||||
\subexc{}
|
||||
\begin{gather*}
|
||||
yz + wx + z +[wz(xy + wz)] \\
|
||||
(yz + z) + wx + (xywz + wz) \\
|
||||
(z + xywz + wz) + wx \\
|
||||
z + wx
|
||||
\end{gather*}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
|
||||
\exc{}{}
|
||||
|
||||
Base case
|
||||
\begin{align*}
|
||||
\sum^{1}_{i=0}i^2 &= \frac{1 \cdot (1 + 1)(2 \cdot 1 + 1)}{6} \\[2ex]
|
||||
1^2 &=\frac{2 \cdot 3}{6} \\[2ex]
|
||||
1 &=\frac{6}{6} \\[2ex]
|
||||
1 &= 1
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
\[ \sum^{k}_{i=0}i^2 = \frac{k (k + 1)(2k + 1)}{6} \]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{k+1}_{i=0}i^2 &= 0^2 + 1^2 + 2^2 + \ldots + k^2 + (k + 1)^2 \\[2ex]
|
||||
&= \frac{k (k + 1)(2k + 1)}{6} + (k + 1)^2 \\[2ex]
|
||||
&= \frac{k (k + 1)(2k + 1) + 6(k + 1)^2}{6} \\[2ex]
|
||||
&= \frac{ (k + 1)(k (2k + 1) + 6(k + 1))}{6} \\[2ex]
|
||||
&= \frac{ (k + 1)(2k^2 + k + 6k + 6)}{6} \\[2ex]
|
||||
&= \frac{ (k + 1)(2k^2 + 7k + 6)}{6} \\[2ex]
|
||||
&= \frac{ (k + 1)(k + 2)(2k + 3)}{6} \\[2ex]
|
||||
&= \frac{(k + 1) ((k + 1) + 1)(2(k + 1) + 1)}{6}
|
||||
\end{align*}
|
||||
|
||||
\qed
|
||||
|
||||
\exc{}
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
\begin{align*}
|
||||
S(0) &= 2^{-0} = 1 \\
|
||||
S(1) &= 2^{-0} + 2^{-1} = 1.5 \\
|
||||
S(2) &= 2^{-0} + 2^{-1} + 2^{-2} = 1.75 \\
|
||||
S(3) &= 2^{-0} + 2^{-1} + 2^{-2} + 2^{-3} = 1.875
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
Based on the results from a, I conjecture that
|
||||
|
||||
\[ S(n) = 2 - 2^{-n} \]
|
||||
|
||||
\subexc{}
|
||||
|
||||
Base case
|
||||
\begin{align*}
|
||||
\sum^{0}_{i=0}2^{-i} &= 2-2^{-0} \\
|
||||
2^{-0} &= 2-1 \\
|
||||
1 &= 1
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
\[ \sum^{n}_{i=0}2^{-i} = 2-2^{-n} \]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{n+1}_{i=0}2^{-i} &= 2^{-0} + 2^{-1} + 2^{-2} + \ldots + 2^{-n} + 2^{-(n+1)} \\
|
||||
&= 2-2^{-n} + 2^{-(n+1)} \\
|
||||
&= 2-2^{-n} + 2^{-n-1} \\
|
||||
&= 2-2^{-n} + 2^{-n}2^{-1} \\
|
||||
&= 2-2^{-n}(1-2^{-1}) \\
|
||||
&= 2-2^{-n}(\frac{2}{2}-\frac{1}{2}) \\
|
||||
&= 2-2^{-n}(\frac{1}{2}) \\
|
||||
&= 2-2^{-n}(2^{-1}) \\
|
||||
&= 2-2^{-n-1} \\
|
||||
&= 2-2^{-(n+1)}
|
||||
\end{align*}
|
||||
|
||||
\qed
|
||||
|
||||
\subexc{}
|
||||
|
||||
\begin{align*}
|
||||
S(n) &> \epsilon \\
|
||||
2-2^{-n} &> \epsilon \\
|
||||
2^{-n} &> \epsilon - 2 \\
|
||||
-n &> \log_2(\epsilon - 2) \\
|
||||
n &< -\log_2(\epsilon - 2) \\
|
||||
\end{align*}
|
||||
|
||||
Assuming $S(n)$ never can reach n,
|
||||
for $S(n)$ to be within $\epsilon$ of $2$, n has to be less than $-\log_2(\epsilon - 2)$
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
|
||||
\exc{}
|
||||
|
||||
Base case
|
||||
\begin{align*}
|
||||
\sum^{1}_{i=1}2^{i-1} \cdot i &= 2^n \cdot (n-1) + 1 \\
|
||||
2^{1-1} \cdot 1 &= 2^1 \cdot (1-1) + 1 \\
|
||||
2^0 \cdot 1 &= 2 \cdot 0 + 1 \\
|
||||
1 \cdot 1 &= 1 \\
|
||||
1 &= 1
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
\[ \sum^{n}_{i=1}2^{i-1} \cdot i = 2^n \cdot (n-1) + 1 \]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{n+1}_{i=1}2^{i-1} \cdot i &= (2^{1-1} \cdot 1) + (2^{2-1} \cdot 2) + \ldots
|
||||
+ (2^{n-1} \cdot n) + (2^{(n+1)-1} \cdot (n+1)) \\
|
||||
&= 2^n \cdot (n-1) + 1 + (2^{(n+1)-1} \cdot (n+1)) \\
|
||||
&= 2^n \cdot (n-1) + 1 + 2^{n} \cdot (n+1) \\
|
||||
&= (2^n \cdot n - 2^n) + 1 + (2^{n} \cdot n + 2^{n}) \\
|
||||
&= 2^n \cdot n - 2^n + 1 + 2^{n} \cdot n + 2^{n} \\
|
||||
&= 2(2^n \cdot n) - 2^n + 2^{n}+ 1 \\
|
||||
&= (4^n \cdot n) + 1 \\
|
||||
&= (2^{n+1} \cdot ((n+1)-1)) + 1
|
||||
\end{align*}
|
||||
|
||||
\qed
|
||||
|
||||
\end{excs}
|
||||
|
||||
\end{document}
|
|
@ -0,0 +1,240 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
\usepackage{ntnu-code}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 5}
|
||||
|
||||
\usepackage{amsthm}
|
||||
\usepackage{mathabx}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
\exc{}
|
||||
|
||||
Base case:
|
||||
\begin{align*}
|
||||
\sum^{m}_{j=1} \frac{1}{j(j+2)} &= \frac{m(3m+5)}{4(m+1)(m+2)} \\[2ex]
|
||||
\frac{1}{1(1+2)} &= \frac{1(3\cdot1 + 5)}{4(1+1)(1+2)} \\[2ex]
|
||||
\frac{1}{3} &= \frac{8}{4(2)(3)} \\[2ex]
|
||||
\frac{1}{3} &= \frac{8}{8(3)} \\[2ex]
|
||||
\frac{1}{3} &= \frac{1}{3}
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
\[ \sum^{m}_{j=1} \frac{1}{j(j+2)} = \frac{m(3m+5)}{4(m+1)(m+2)} \]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{m+1}_{j=1} \frac{1}{j(j+2)}
|
||||
&= \frac{1}{1(1+2)} + \frac{1}{2(2+2)} + \ldots + \frac{1}{m(m+2)} + \frac{1}{(m+1)((m+1)+2)} \\[2ex]
|
||||
&= \frac{m(3m+5)}{4(m+1)(m+2)} + \frac{1}{(m+1)((m+1)+2)} \\[2ex]
|
||||
&= \frac{3m^2+5m}{4m^2+12m+8} + \frac{1}{m^2+4m+3} \\[2ex]
|
||||
&= \frac{(m^2+4m+3)(3m^2+5m) + 4m^2+12m+8}{(m^2+4m+3)(4m^2+12m+8)} \\[2ex]
|
||||
&= \frac{3m^4+17m^3+29m^2+15m + 4m^2+12m+8}{(m^2+4m+3)(4m^2+12m+8)} \\[2ex]
|
||||
&= \frac{3m^4+17m^3+33m^2+27m+8}{(m^2+4m+3)(4m^2+12m+8)} \\[4ex]
|
||||
&\text{(Here, I used a calculator to factorize the expression)} \\[4ex]
|
||||
&= \frac{3m^2+11m+8}{4m^2+20m+24} \\[2ex]
|
||||
&= \frac{(m+1)(3m+8)}{4(m^2+5m+6)} \\[2ex]
|
||||
&= \frac{(m+1)(3(m+1)+5)}{4(m+2)(m+3)} \\[2ex]
|
||||
&= \frac{(m+1)(3(m+1)+5)}{4((m+1)+1)((m+1)+2)} \\[2ex]
|
||||
\end{align*}
|
||||
\qed
|
||||
|
||||
\exc{}
|
||||
|
||||
\begin{align*}
|
||||
a_{m+1} &= 2^{2(m + 1) + 1} + 1 \\
|
||||
&= 2^{2m+3} + 1 \\
|
||||
&= 2^{2m+1}2^2 + 1 \\
|
||||
&= (2^{2m+1} + 1)2^2 - (1)2^2 + 1 \\
|
||||
&= 4a_{m} - 4 + 1 \\
|
||||
&= 4a_{m} - 3 \\
|
||||
\end{align*}
|
||||
|
||||
$(a_m \bmod 3 = 0) \wedge (-3 \bmod 3 = 0) \Rightarrow ((4a_m - 3) \bmod 3 = 0)$
|
||||
|
||||
\exc{}
|
||||
|
||||
Base case:
|
||||
\begin{align*}
|
||||
\sum^m_{i=1} iL_i &= mL_{m+2} - L_{m+3} + 4 \\
|
||||
1 \cdot L_1 &= 1 \cdot L_{1+2} - L_{1+3} + 4 \\
|
||||
1 &= 4 - 7 + 4 \\
|
||||
1 &= 1 \\
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
\[ \sum^m_{i=1} i L_i = mL_{m+2} - L_{m+3} + 4 \]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{m+1}_{i=1} iL_i &= 1L_1 + 2L_2 + \ldots + mL_m + (m+1)L_{m+1} \\
|
||||
&= mL_{m+2} - L_{m+3} + 4 + (m+1)L_{m+1} \\
|
||||
&= mL_{m+2} - L_{m+3} + 4 + mL_{m+1} + L_{m+1} \\
|
||||
&= m(L_{m+2} + L_{m+1}) - L_{m+3} + L_{m+1} + 4 \\
|
||||
&= mL_{m+3} - L_{m+3} + L_{m+1} + 4 \\
|
||||
&= (m+1)L_{m+3} - L_{m+3} - L_{m+3} + L_{m+1} + 4 \\
|
||||
&= (m+1)L_{m+3} - L_{m+3} - L_{m+2} - L{m+1} + L_{m+1} + 4 \\
|
||||
&= (m+1)L_{m+3} - L_{m+3} - L_{m+2} + 4 \\
|
||||
&= (m+1)L_{m+3} - L_{m+4} + 4 \\
|
||||
&= (m+1)L_{(m+1)+2} - L_{(m+1)+3} + 4 \\
|
||||
\end{align*}
|
||||
\qed
|
||||
|
||||
\exc{}
|
||||
|
||||
Base case:
|
||||
\begin{align*}
|
||||
\sum^1_{i=1}(-1)^{i+1}i^2 &= (-1)^{1+1}\sum^{1}_{i=1}i \\
|
||||
(-1)^{1+1} \cdot 1^2 &= (-1)^{1+1} \cdot 1 \\
|
||||
(-1)^{2} &= (-1)^{2} \\
|
||||
1 &= 1 \\
|
||||
\end{align*}
|
||||
|
||||
Assume:
|
||||
|
||||
\[ \sum^m_{i=1}(-1)^{i+1}i^2 = (-1)^{m+1}\sum^{m}_{i=1}i = (-1)^{m+1}\left(\frac{m(m+1)}{2}\right)\]
|
||||
|
||||
\begin{align*}
|
||||
\sum^{m+1}_{i=1} &= (-1)^{1+1} \cdot 1^2 + (-1)^{2+1} \cdot 2^2 + \ldots + (-1)^{m+1} m^2 + (-1)^{(m+1)+1} (m+1)^2 \\
|
||||
&= (-1)^{m+1}\frac{m(m+1)}{2} + (-1)^{(m+1)+1} (m+1)^2 \\
|
||||
&= (-1)^{m+1} \left(\frac{m(m+1)}{2} + (-1) (m+1)^2 \right) \\
|
||||
&= (-1)^{m+1} (m+1) \left( \frac{m}{2} - (m+1) \right) \\
|
||||
&= (-1)^{m+1} (-1) (m+1) \left( -\frac{m}{2} + (m+1) \right) \\
|
||||
&= (-1)^{m+2} (m+1) \left( -\frac{m}{2} + (m+1) \right) \\
|
||||
&= (-1)^{m+2} (m+1) \left( \frac{-m + 2(m+1)}{2} \right) \\
|
||||
&= (-1)^{m+2} (m+1) \left( \frac{-m + 2m + 2}{2} \right) \\
|
||||
&= (-1)^{m+2} (m+1) \left( \frac{m + 2}{2} \right) \\
|
||||
&= (-1)^{m+2} \left( \frac{(m+1)(m + 2)}{2} \right) \\
|
||||
&= (-1)^{(m+1)+1} \left( \frac{(m+1)((m+1) + 1)}{2} \right) \\
|
||||
\end{align*}
|
||||
\qed
|
||||
|
||||
\exc{}
|
||||
|
||||
\begin{subexcs}
|
||||
\subexc{}
|
||||
|
||||
$R$ is reflexive because $x \bmod x = 0$
|
||||
|
||||
$R$ is not symmetric because $ 2 \bmod 1 = 0$ but $1 \bmod 2 = 1$
|
||||
|
||||
$R$ is transitive because
|
||||
\begin{align*}
|
||||
xRy &\Leftrightarrow (y = nx), &&n \in \mathbb{Z} \\
|
||||
yRz &\Leftrightarrow (z = my = m(nx)), &&m \in \mathbb{Z} \\
|
||||
z=nmx &\Leftrightarrow (z \bmod x = 0)
|
||||
\end{align*}
|
||||
|
||||
\subexc{}
|
||||
|
||||
$R$ is reflexive because \[ A \cap C = A \cap C \]
|
||||
|
||||
$R$ is symmetric because \[ A \cap C = B \cap C \Leftrightarrow B \cap C = A \cap C \]
|
||||
|
||||
$R$ is transitive because \[ (A \cap C = B \cap C) \wedge (B \cap C = D \cap C) \Rightarrow A \cap C = D \cap C \]
|
||||
|
||||
\subexc{}
|
||||
|
||||
$R$ is not reflexive because \[ l_1 \not\perp l_1 \]
|
||||
|
||||
$R$ is symmetric because \[ l_1 \perp l_2 \Leftrightarrow l_2 \perp l_1 \]
|
||||
|
||||
$R$ is not transitive because \[ l_1 \perp l_2 \wedge l_2 \perp l_3 \Rightarrow l_1 \not\perp l_3 \]
|
||||
|
||||
\subexc{}
|
||||
|
||||
$R$ is not reflexive because \[ (2n + 1) + (2n + 1) = 2(2n + 1) = 2k \]
|
||||
|
||||
$R$ is symmetric because \[ x+y = y+x = 2n+1 \]
|
||||
|
||||
$R$ is not transitive because an odd number can only be the sum of two integers if one is odd and the other is even
|
||||
|
||||
Case 1: $x$ is even and $y$ is odd:
|
||||
\begin{align*}
|
||||
x + y &= 2n+1 \\
|
||||
y + z &= 2n+1 \Rightarrow z = 2k \\
|
||||
x+z &= 2k_1 + 2k_2 = 2(k_1 + k_2) = 2k
|
||||
\end{align*}
|
||||
|
||||
Case 2: $x$ is odd and $y$ is even:
|
||||
\begin{align*}
|
||||
x + y &= 2n+1 \\
|
||||
y + z &= 2n+1 \Rightarrow z = 2k + 1 \\
|
||||
x+z &= (2k_1+1) + (2k_2+1) = 2(k_1 + k_2) + 2 = 2(k_1 + k_2 + 1) = 2k
|
||||
\end{align*}
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
|
||||
In order to show that this is an equivalence relation, the relation has to be reflexive, symmetric and transitive
|
||||
|
||||
Reflexive:
|
||||
\[ ab = ba \]
|
||||
|
||||
Symmetric:
|
||||
\[ (ad = bc) \Leftrightarrow (bc = ad) \]
|
||||
|
||||
Transitive:
|
||||
\[ (ad = bc) \wedge (cf = de) \Leftrightarrow \left(\frac{a}{b} = \frac{c}{d}\right) \wedge \left(\frac{c}{d} = \frac{e}{f}\right) \Rightarrow \left(\frac{a}{b} = \frac{e}{f}\right) \Leftrightarrow (af = be) \]
|
||||
|
||||
\exc{}
|
||||
|
||||
In order to show that this is an equivalence relation, the relation has to be reflexive, symmetric and transitive
|
||||
|
||||
Reflexive:
|
||||
\[ x+y = x+y \]
|
||||
|
||||
Symmetric:
|
||||
\[ (x+y = u+v) \Leftrightarrow (u+v = x+y) \]
|
||||
|
||||
Transitive:
|
||||
\[ (x+y = u+v) \wedge (u+v = m+n) \Rightarrow (x+y = m+n) \]
|
||||
|
||||
In this case, you could either use the fact that there are only specific integers that will sum to another integer, or check every relation between every tuple in order to calculate the equivalence classes. I decided to solve it by automating the process.
|
||||
|
||||
\codeFile{scripts/ex7.hs}{haskell}
|
||||
|
||||
Output:
|
||||
\begin{verbatim}
|
||||
[(1,3),(2,2),(3,1)]
|
||||
[(1,5),(2,4),(3,3),(4,2),(5,1)]
|
||||
[(1,1)]
|
||||
\end{verbatim}
|
||||
|
||||
therefore
|
||||
|
||||
\begin{align*}
|
||||
[(1,3)] &= \{(1,3), (2,2), (3,1)\} \\
|
||||
[(2,4)] &= \{(1,5), (2,4), (3,5), (4,2), (5,1)\} \\
|
||||
[(1,1)] &= \{(1,1)\}
|
||||
\end{align*}
|
||||
|
||||
\exc{}
|
||||
In order to show that this is an equivalence relation, the relation has to be reflexive, symmetric and transitive
|
||||
|
||||
Reflexive:
|
||||
\[ x - x \bmod 3 = 0 \bmod 3 = 0 \]
|
||||
|
||||
Symmetric:
|
||||
\[ (x - y \bmod 3 = 0) \Leftrightarrow (x,y= 3k_1 + r, 3k_2 + r) \Leftrightarrow (y - x \bmod 3 = 3(k_2 - k_1) + r - r = 0) \]
|
||||
|
||||
Transitive:
|
||||
\[ (ad = bc) \wedge (cf = de) \Leftrightarrow \left(\frac{a}{b} = \frac{c}{d}\right) \wedge \left(\frac{c}{d} = \frac{e}{f}\right) \Rightarrow \left(\frac{a}{b} = \frac{e}{f}\right) \Leftrightarrow (af = be) \]
|
||||
|
||||
The equivalence classes will contain the numbers that has the same remainder after dividing by $3$, since subtracting them from each other will remove the remainder and make the number divisible by $3$.
|
||||
|
||||
therefore the partition of $A$ induced by $R$ will be
|
||||
|
||||
\[ \{\{1,4,7\}, \{2,5\}, \{3,6\}\} \]
|
||||
|
||||
|
||||
\end{excs}
|
||||
|
||||
|
||||
|
||||
\end{document}
|
|
@ -0,0 +1,30 @@
|
|||
import Control.Monad (mapM_)
|
||||
import Data.List (nub)
|
||||
|
||||
type Pair = (Integer, Integer)
|
||||
type Relation = (Pair, Pair)
|
||||
|
||||
cartesianProduct :: [Integer] -> [Pair]
|
||||
cartesianProduct domain = [ (x,y) | x <- domain,
|
||||
y <- domain ]
|
||||
|
||||
calculateRelations :: [Pair] -> (Pair -> Pair -> Bool) -> [Relation]
|
||||
calculateRelations set relation
|
||||
= nub [ (p1, p2) | p1 <- set,
|
||||
p2 <- set,
|
||||
relation p1 p2 ]
|
||||
|
||||
getRelatedPairsOf :: Pair -> [Relation] -> [Pair]
|
||||
getRelatedPairsOf s r = [ p2 | (p1, p2) <- r,
|
||||
p1 == s ]
|
||||
|
||||
main :: IO ()
|
||||
main = do
|
||||
let
|
||||
-- set a be the cartesian product of two lists of integers from 1 to and including 5
|
||||
setA = cartesianProduct [1..5]
|
||||
-- set r be the relation on a that satisfy the following condition
|
||||
r = calculateRelations setA (\(x,y) (u,v) -> x+y == u+v)
|
||||
|
||||
-- filter out the equivalence classes of the following pairs
|
||||
mapM_ (print . flip getRelatedPairsOf r) [(1,3), (2,4), (1,1)]
|
|
@ -0,0 +1,27 @@
|
|||
import Control.Monad (mapM_)
|
||||
import Data.List (partition)
|
||||
|
||||
-- | Split off any nums that satisfy the relation and return the split and the rest in a tuple
|
||||
splitOffRelatedNums :: (Integer -> Integer -> Bool) -> [Integer] -> ([Integer],[Integer])
|
||||
splitOffRelatedNums relation nums = partition (relation x) nums
|
||||
where
|
||||
x = head nums
|
||||
|
||||
-- | Split off equivalence groups until there are no more that satisfy the condition
|
||||
getEquivalenceClasses :: (Integer -> Integer -> Bool) -> [Integer] -> [[Integer]]
|
||||
getEquivalenceClasses relation nums
|
||||
= case nums of
|
||||
[] -> []
|
||||
nums -> x : getEquivalenceClasses relation xs
|
||||
where
|
||||
(x,xs) = splitOffRelatedNums relation nums
|
||||
|
||||
main :: IO ()
|
||||
main = do
|
||||
let
|
||||
setA = [1..7]
|
||||
|
||||
relation :: Integer -> Integer -> Bool
|
||||
relation x y = (x - y) `mod` 3 == 0
|
||||
|
||||
mapM_ print $ getEquivalenceClasses relation setA
|
|
@ -0,0 +1,231 @@
|
|||
\documentclass[12pt]{article}
|
||||
\usepackage{ntnu}
|
||||
\usepackage{ntnu-math}
|
||||
\usepackage{ntnu-code}
|
||||
|
||||
\author{Øystein Tveit}
|
||||
\title{MA0301 Exercise 6}
|
||||
|
||||
\usepackage{amsthm}
|
||||
\usepackage{mathabx}
|
||||
|
||||
\begin{document}
|
||||
\ntnuTitle{}
|
||||
\break{}
|
||||
|
||||
\begin{excs}
|
||||
\exc{}
|
||||
|
||||
\begin{subexcs}
|
||||
|
||||
\subexc{}
|
||||
|
||||
In order to show that this is a partial order, the relation has to be reflexive, antisymmetric and transitive
|
||||
|
||||
However, it is not antisymmetric because
|
||||
|
||||
\[ 4-2 \bmod 2 = 0 \wedge 2 - 4 \bmod 2 = 0 \]
|
||||
|
||||
\subexc{}
|
||||
|
||||
In order to show that this is a partial order, the relation has to be reflexive, antisymmetric and transitive
|
||||
|
||||
However, it is not antisymmetric because
|
||||
|
||||
\[ (1,2)R(1,3) \wedge (1,3)R(1,2) \wedge (1,2) \neq (1,3) \]
|
||||
|
||||
\end{subexcs}
|
||||
|
||||
\exc{}
|
||||
|
||||
\begin{figure}[H]
|
||||
\begin{mgraphbox}[width=5cm]
|
||||
\center
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\tikzset{every node/.style={shape=circle,draw,inner sep=2pt}}
|
||||
|
||||
\node (a) at (0,0) {$1$};
|
||||
\node (b) at (1,1) {$2$};
|
||||
\node (c) at (-1,1) {$3$};
|
||||
\node (d) at (0,2) {$6$};
|
||||
\node (e) at (-2,2) {$9$};
|
||||
\node (f) at (-1,3) {$18$};
|
||||
|
||||
\draw (a) -- (b) -- (d) -- (f) -- (e) -- (c) -- (a);
|
||||
\draw (c) -- (d);
|
||||
\end{tikzpicture}
|
||||
\end{mgraphbox}
|
||||
\caption{Hasse diagram of $R$}
|
||||
\end{figure}
|
||||
|
||||
\exc{}
|
||||
|
||||
\begin{subexcs}
|
||||
|
||||
\subexc{}
|
||||
|
||||
In order to show that this is a partial order, the relation has to be reflexive, antisymmetric and transitive
|
||||
|
||||
Reflexive:
|
||||
|
||||