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\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 $x=b \land y=b$. $M\nmodels\varphi$.