\item \self Compute the propositional formula of the following circuit.

\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 (8,-1) {$z$};

    \node (bm) at (3,-1) {};
    \node (cm1) at (2,-2) {};
    \node (cm2) at (3,-2) {};
    \node (bc_c1) at (2,-1.1) {};
    \node (bc_c2) at (2.1,-1) {};
    \node (bc_c3) at (2,-.9) {};
    \node (not1m) at (3,0) {};
    \node (and1m) at (5,-.5) {};
    \node (nand1m) at (5,-1.5) {};

    \node[not gate US, draw, logic gate inputs=n] at ($(2,0)$) (not1) {\tiny NOT};
    \node[and gate US, draw, logic gate inputs=nn] at ($(4,-.5)$) (and1) {\tiny AND};
    \node[nand gate US, draw, logic gate inputs=nn] at ($(4,-1.5)$) (nand1) {\tiny NAND};
    \node[xor gate US, draw, logic gate inputs=nn] at ($(6,-1)$) (xor1) {\tiny XOR};

    \draw (a.east) -- (not1.input);
    \draw (not1.output) -| (not1m.center) node[above] {$w$} |- (and1.input 1);
    \draw (c.east) -| (cm1.center) node[circle, fill,inner sep=2pt] {} -- (bc_c1.center) -- (bc_c2.center) -- (bc_c3.center) |- (and1.input 2);
    \draw (b.east) -| (bm.center) |- (nand1.input 1);
    \draw (c.east) -| (cm2.center) |- (nand1.input 2);
    \draw (and1.output) -| (and1m.center) node[above] {$x$} |- (xor1.input 1);
    \draw (nand1.output) -| (nand1m.center) node[below] {$y$} |- (xor1.input 2);
    \draw (xor1.output) -- (z);
\end{tikzpicture}