Questionnaire-2024/equivalence_checking/0001_sol.tex

\tikzstyle{branch}=[fill,shape=circle,minimum size=3pt,inner sep=0pt]
\begin{tikzpicture}[label distance=2mm]
    \node (a) at (0,0) {$a$};
    \node (b) at (0,-1) {$b$};
    \node (c) at (0,-2) {$c$};
    \node (z) at (6,-1) {$z$};

    \node (am) at (1,0) {};
    \node (bm) at (1,-1) {};

    \node[and gate US, draw, logic gate inputs=nn] at ($(2,-.5)$) (and1) {\tiny AND};
    \node[or gate US, draw, logic gate inputs=nn] at ($(4,-1)$) (or1) {\tiny OR};

    \node (and1m) at (3,-.5) {};
    \node (c1m) at (3,-2) {};

    \draw (a.east) -| (am.center) |- (and1.input 1);
    \draw (b.east) |- (bm.center) |- (and1.input 2);
    \draw (and1.output) |-  (and1m.center) node[above] {$y$} |- (or1.input 1);
    \draw (c.east) |- (c1m.center) |- (or1.input 2);
    \draw (or1.output) -- (z);
\end{tikzpicture}

The inputs are denoted by $a$, $b$, and $c$ and the output is denoted by $z$.
We assign temporary variable names to the inner wires; in this case we use $y$.
Using these variables, we can create the propositional formula over the inputs and the output.
%
\begin{equation*} 
\begin{split}
z & = y \lor c \\
 & = (a \land b) \lor c
\end{split}
\end{equation*}