Answer
~(PvQ)
see figure.
Work Step by Step
1. Use Theorem 2.1.1 to simplify the given expression:
(~P^~Q)v(~P^Q)v(P^~Q)$\equiv$~P^(~QvQ)v(P^~Q)$\equiv$
~P^(t)v(P^~Q)$\equiv$~Pv(P^~Q)$\equiv$(~PvP)^(~Pv~Q)$\equiv$
t^(~Pv~Q)$\equiv$~Pv~Q$\equiv$~(P^Q)
2. The above can be presented with a circuit of two gates as shown.
