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\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*}