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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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    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)2012-05-07

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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 t$h$at I didn't. Thanks for the quick catch.2012-05-07
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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.