How do I prove the following statement is a tautology, without using truth tables? $$[¬P ∧ (P ∨ Q)] → Q$$
I know that if we assume $Q ≡ T$ then no matter what the truth value of what is to the left of the implication operator is, the statement will be a tautology. But if we assume that $Q ≡ F$ then there could be two possibilities of the outcome of the statement: If $\;[¬P ∧ (P ∨ Q)] ≡ T,\;$ then the statement is false, and if $\;[¬P ∧ (P ∨ Q)] ≡ F,\;$ then the statement is true (according to the truth table of implication statements: $\;T → F = F\;\text{ and }\;F → F = T.)$
Is there a way of proving $\;[¬P ∧ (P ∨ Q)]\rightarrow Q\;$ is always true without using any truth tables, instead can it be solely proven by words/logic? Or am I just being dumb?
Best Answer
As you note, if $Q$ is true, then the implication is true.
And if $Q$ is false, we have that $\lnot P \land (P \lor Q) \equiv \underbrace{\underbrace{(\lnot P \land P)}_{F} \lor \underbrace{(\lnot P \land \underbrace{Q}_{F})}_{F}}_{F}$ and any implication with a false premise is true.
Hence, the implication is a tautology.