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

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.

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:

$$1). (p ∧ q) ⇒ r - premise$$

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

$$3). p - assumption$$

$$4). q - assumption$$

$$5)....$$

• I have actually solved the question, thank you for your insight. Oct 5, 2020 at 2:05