Natural Deduction Proof for $p\land q \rightarrow r \vdash (p \rightarrow r) \lor (q \rightarrow r)$

Question

Can anyone give me some hints on how to prove $p\land q \rightarrow r \vdash (p \rightarrow r) \lor (q \rightarrow r)$ with natural deduction? I have spend hours trying to prove it to no avail. I know that double negation, negation rules are necessary here, however I kept get stuck and redoing it all over again many times.

Any kind of hints would be greatly appreciated, thank you.

Answer 1

Since double negation is needed a good aproach is to assume the negation of the conclusion as your first step, and then try to derivate a contradiction so that way you use $I¬$ and after that you use $E¬¬$. Try with this assumtions:

$1). (p ∧ q) ⇒ r - premise$

$2). ¬((p ⇒ r) ∨ (q ⇒ r)) - assumption$

$3). p - assumption$

$4). q - assumption$

$5)....$