0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Sep 6 '16 at 9:45
  • $\begingroup$ I think the OP is remembering this incorrectly. Or else it is a trick question. $\endgroup$ – Christopher Carl Heckman Sep 11 '16 at 0:02
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.