# Logic and Proving using rules of inference.

We recently had our exam and there's this one problem I couldn't finish. I think I'm missing something obvious because I'm really close to the proof.

The queston is this:

Given statements $p$, $q$ and $r$, prove the series of hypothesis
$(p \vee q) \implies \neg\ r$
$s \implies \neg\ p$
$s \wedge \neg\ q$
logically implies the conclusion, $r$

My proof:
$s \wedge \neg\ q$
$s$
$s \implies \neg\ p$
$\neg\ p$
$\neg\ q$
$\neg\ p \wedge \neg\ q$
$(p \vee q) \implies \neg\ r$
$\neg\ (p \vee q)$

I wasn't able to arrive at $r$ but I think the last two steps can lead to $r$ with a little trick that I'm not aware of. I would write the rule of inference that I used at each step but posting here is a little different than other sites that I'm used to.

• I'm not sure this is correct. If the conclusion is $r$, there's nothing in the list that implies $r$. In fact, it's possible that $\sim r$ could be always true. – астон вілла олоф мэллбэрг Sep 6 '16 at 9:45
• I think the OP is remembering this incorrectly. Or else it is a trick question. – Christopher Carl Heckman Sep 11 '16 at 0:02

If the question was set correctly, it is a trick question purposely to identify students who go and prove an invalid theorem. To see why it is invalid it suffices to construct a situation (model) that satisfies the premises but not the conclusion. One such situation is when $p,q,r$ are false and $s$ is true.