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