# Prove $( p \land \lnot q ) \lor ( p \land q) \Leftrightarrow p$

I'm trying to prove $( p \land \lnot q ) \lor ( p \land q) \Leftrightarrow p$

by doing the following:

\begin{align} ( p \lor p ) \land \lnot &q & Distributive \\ p \land \lnot &q & Idempotent \\ \end{align}

$q$ is still $\{T,F\}$, as is $\lnot q$. Therefore, $( p \land \lnot q ) \lor ( p \land q)$ is contingent upon both q and p.

But my book says it should reduce to p.

Does anyone out there have any insights?

• You applied distributivity incorrectly! $(a \wedge b) \vee (a \wedge c) = a \wedge (b \vee c)$ – Lynn May 15 '16 at 22:28
• @Lynn - Thanks! – StudentsTea May 15 '16 at 22:32

Note that the correct distributive statement should be:$$p \land (\lnot q \lor q)$$ The final result should follow quite easily from this.
\begin{align} ( p \land \lnot q ) &\lor ( p \land q ) \\ p \land ( \lnot q &\lor q ) &Distributive \\ p \land &T &Tautology \\ &p &Identity\\ \end{align}
$(p \wedge \lnot q) \vee (p \wedge q) \iff p \wedge (q \vee \lnot q)$
Since $(q \vee \lnot q)$ is always true this gives $p$.