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82 lines
1.8 KiB
82 lines
1.8 KiB
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\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
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\Tree
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[.$\exists x$
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[.$\forall y$
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[.$\imp$
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[.$P$ $x$ $y$ ]
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[.$\lor$
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[.$Q$ $x$ $y$ ]
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[.$R$ $x$ $y$ ]
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]
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]
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]
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]
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\end{tikzpicture}
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\begin{multicols}{2}
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$$x=a \land y=a$$
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\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
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\Tree
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% [.$\exists x$
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% [.$\forall y$
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[.$\imp$
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[.$P$ $a$ $a$ ]
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[.$\lor$
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[.$Q$ $a$ $a$ ]
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[.$R$ $a$ $a$ ]
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]
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]
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% ]
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% ]
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\end{tikzpicture}
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\begin{align*}
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&P(a,a) \imp (Q(a,a) \lor R(a,a)) = \\
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&\true \imp (\true \lor \true) = \true
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\end{align*}
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\columnbreak
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$$x=a \land y=b$$
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\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
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\Tree
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% [.$\exists x$
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% [.$\forall y$
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[.$\imp$
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[.$P$ $a$ $b$ ]
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[.$\lor$
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[.$Q$ $a$ $b$ ]
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[.$R$ $a$ $b$ ]
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]
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]
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% ]
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% ]
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\end{tikzpicture}
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\begin{align*}
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&P(a,b) \imp (Q(a,b) \lor R(a,b)) = \\
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&\true \imp (\true \lor \false) = \true
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\end{align*}
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\end{multicols}
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\begin{centering}
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$$x=a \land y=c$$
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\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
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\Tree
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% [.$\exists x$
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% [.$\forall y$
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[.$\imp$
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[.$P$ $a$ $c$ ]
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[.$\lor$
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[.$Q$ $a$ $c$ ]
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[.$R$ $a$ $c$ ]
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]
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]
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% ]
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% ]
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\end{tikzpicture}
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\begin{align*}
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&P(a,c) \imp (Q(a,c) \lor R(a,c)) = \\
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&\false \imp (\true \lor \true) = \true
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\end{align*}
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\end{centering}
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We have found an $x$, such that for all $y$ the formula evaluates to \textit{true}. Therefore $$M \models \varphi.$$
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