Questionnaire-2024/predicate_logic/2001_sol.tex

\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
\Tree
 [.$\exists x$
    [.$\forall y$
        [.$\imp$
            [.$P$ $x$ $y$ ]
            [.$\lor$
                [.$Q$ $x$ $y$ ]
                [.$R$ $x$ $y$ ]
            ]
        ]
    ]
 ]
\end{tikzpicture}


\begin{multicols}{2}

  $$x=a \land y=a$$
\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
\Tree
% [.$\exists x$
%    [.$\forall y$
        [.$\imp$
            [.$P$ $a$ $a$ ]
            [.$\lor$
                [.$Q$ $a$ $a$ ]
                [.$R$ $a$ $a$ ]
            ]
        ]
%    ]
% ]
\end{tikzpicture}
 \begin{align*}
   &P(a,a) \imp (Q(a,a) \lor R(a,a)) = \\
   &\true \imp (\true \lor \true) = \true
 \end{align*}
\columnbreak
  $$x=a \land y=b$$
\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
\Tree
% [.$\exists x$
%    [.$\forall y$
        [.$\imp$
            [.$P$ $a$ $b$ ]
            [.$\lor$
                [.$Q$ $a$ $b$ ]
                [.$R$ $a$ $b$ ]
            ]
        ]
%    ]
% ]
\end{tikzpicture}
 \begin{align*}
   &P(a,b) \imp (Q(a,b) \lor R(a,b)) = \\
   &\true \imp (\true \lor \false) = \true
 \end{align*}
\end{multicols}
\begin{centering}
  $$x=a \land y=c$$
\begin{tikzpicture}[every tree node/.style={draw,circle},sibling distance=.15cm]
\Tree
% [.$\exists x$
%    [.$\forall y$
        [.$\imp$
            [.$P$ $a$ $c$ ]
            [.$\lor$
                [.$Q$ $a$ $c$ ]
                [.$R$ $a$ $c$ ]
            ]
        ]
%    ]
% ]
\end{tikzpicture}
 \begin{align*}
   &P(a,c) \imp (Q(a,c) \lor R(a,c)) = \\
   &\false \imp (\true \lor \true) = \true
 \end{align*}
\end{centering}

We have found an $x$, such that for all $y$ the formula evaluates to \textit{true}. Therefore $$M \models \varphi.$$