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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?

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    $\begingroup$ You applied distributivity incorrectly! $(a \wedge b) \vee (a \wedge c) = a \wedge (b \vee c)$ $\endgroup$ – Lynn May 15 '16 at 22:28
  • $\begingroup$ @Lynn - Thanks! $\endgroup$ – StudentsTea May 15 '16 at 22:32
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Note that the correct distributive statement should be:$$p \land (\lnot q \lor q)$$ The final result should follow quite easily from this.

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As Lynn correctly pointed out, I applied the Distributive Law incorrectly. It should go:

$$\begin{align} ( p \land \lnot q ) &\lor ( p \land q ) \\ p \land ( \lnot q &\lor q ) &Distributive \\ p \land &T &Tautology \\ &p &Identity\\ \end{align}$$

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Using distributivity we have :

$(p \wedge \lnot q) \vee (p \wedge q) \iff p \wedge (q \vee \lnot q)$

Since $(q \vee \lnot q)$ is always true this gives $p$.

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