So I have to prove [(p ∨ q) ∧ (p → s) ∧ (q → t)] → (s ∨ t) without using truth tables and give the proof technique used.
I am not sure how to do this, I thought that I should use logical equivalences, which I'm not really good at in the first place, so I couldn't really get anywhere after changing conditional statements like (p → s) to (¬ p ∨ s) and trying to distribute between them.
However, they said that I have to give the proof technique, so that would be something among direct proofs, proof by contraposition, proof by contradiction... Right? If that is so, then I am still lost as to what to do.
Can someone please give me an idea? Thank you.
Natural Deduction. Argue from cases, via the Rule of Inference, Disjunctive Elimination: $p\vee q, p\to r, q\to r\vdash r$ .
Take the given premise, which entails $(p\vee q), (p\to s), (q\to t)$. Thus assuming $p$ infers —for some reasons— $s\vee t$. Also assuming $q$ infers —for similar reasons— $s\vee t$. So assuming $p$ or $q$, that is $p\vee q$, will infer $s\vee t$.
Therefore the premise entails $s\vee t$.$$\therefore (p\vee q)\wedge(p\to s)\wedge(q\to t)\vdash (s\vee t)$$
$\blacksquare$