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I cannot see any steps to this problem! Surely the answer is obvious? Is there a particular law which is used to make this statement?

$$((Q \wedge ¬P) \vee (Q \wedge P)) = Q$$

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Shouldn't it be $$((Q \wedge ¬P) \vee (Q \wedge P)) = Q$$? –  Pedro Tamaroff May 7 '12 at 3:03
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It's hard to say, because you haven't said what steps are allowed. For example, if you already know that $\vee$ distributes over $\wedge$, the thing is very simple: you factor out the $Q$ from the disjuncts, obtaining $(Q\wedge(P\vee\lnot P))=Q$, etc. –  MJD May 7 '12 at 3:04
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MarkDominus yes you are right, I can see the answer now. $$(Q \wedge (P \vee ¬P)) = Q$$ (Distributive Law) $$(Q \wedge T) = Q$$ (Excluded Middle) $$Q = Q$$ (Identity) –  Danny Rancher May 7 '12 at 3:06

2 Answers 2

HINT: $$(Q\land\lnot P)\lor(Q\land P)\equiv Q\land(\lnot P\lor P)$$ by distributivity. Or of course you can simply use a truth table, if you're allowed to do so in this problem.

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@Mark: I'm sure that I didn't. Thanks for the quick catch. –  Brian M. Scott May 7 '12 at 3:07

The shape of this makes me think of DeMorgan's Laws. You could start with $Q=Q \vee False$ But you could just do a truth table.

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