\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm] \Tree [.$\exists x$ [.$\forall y$ [.$\lor$ [.$\imp$ [.$P$ $x$ $y$ ] [.$Q$ $x$ $y$ ] ] [.$\imp$ [.$P$ $y$ $x$ ] [.$R$ $x$ $y$ ] ] ] ] ] \end{tikzpicture} \begin{itemize} \item $x=a$ \end{itemize} \begin{multicols}{2} $$x=a \land y=a$$ \begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm] \Tree [.$\lor$ [.$\imp$ [.$P$ $a$ $a$ ] [.$Q$ $a$ $a$ ] ] [.$\imp$ [.$P$ $a$ $a$ ] [.$R$ $a$ $a$ ] ] ] \end{tikzpicture} \begin{align*} &(P(a,a) \imp Q(a,a)) \lor (P(a,a) \imp R(a,a)) = \\ &(\false \imp \true) \lor (\false \imp \true) = \true \end{align*} \columnbreak $$x=a \land y=b$$ \begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm] \Tree [.$\lor$ [.$\imp$ [.$P$ $a$ $b$ ] [.$Q$ $a$ $b$ ] ] [.$\imp$ [.$P$ $b$ $a$ ] [.$R$ $a$ $b$ ] ] ] \end{tikzpicture} \begin{align*} &(P(a,b) \imp Q(a,b)) \lor (P(b,a) \imp R(b,a)) = \\ &(\true \imp \false) \lor (\true \imp \false) = \false \end{align*} \end{multicols} \begin{itemize} \item $x=b$ \end{itemize} \begin{centering} $$x=b \land y=a$$ \begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm] \Tree [.$\lor$ [.$\imp$ [.$P$ $b$ $a$ ] [.$Q$ $b$ $a$ ] ] [.$\imp$ [.$P$ $a$ $b$ ] [.$R$ $b$ $a$ ] ] ] \end{tikzpicture} \begin{align*} &(P(b,a) \imp Q(b,a)) \lor (P(a,b) \imp R(b,a)) = \\ &(\true \imp \false) \lor (\true \imp \false) = \false \end{align*} \end{centering} We do not need to evaluate $x=b \land y=b$. $M\nmodels\varphi$.