120 lines
3.9 KiB
TeX
120 lines
3.9 KiB
TeX
\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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\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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{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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\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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(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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\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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\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{}
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$\exists S(x)[\neg H(x)]$
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\subexc{}
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$\forall S(x)[\neg H(x)]$
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\subexc{}
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$\forall \neg H(x) \exists S(x)$
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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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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{}
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The formula is false because $p(z, y)$ and $\neg p(z, x)$ cannot be fulfilled at the same time. $z \geq 0 \wedge \neg (z \geq 0) \equiv F$
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\setsubexc{4}
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\subexc{}
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\fbox{see comment in ovsys}
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\end{subexcs}
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\exc{}
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\begin{align*}
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\neg (\forall x [p(x) \wedge q(x)]) &\equiv \exists x [\neg(p(x) \wedge q(x))] \\
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&\equiv \exists x [\neg p(x) \vee \neg q(x)]
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\end{align*}
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\exc{}
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\begin{align*}
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\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)]) \\
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&\equiv \forall x \exists y \neg[p(y) \vee \neg q(x,y)] \\
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&\equiv \forall x \exists y [\neg p(y) \wedge \neg\neg q(x,y)] \\
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&\equiv \forall x \exists y [\neg p(y) \wedge q(x,y)]
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\end{align*}
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\end{excs}
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\end{document}
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