1. (r ∧ ¬s) ∨ (q ∧ ¬s)
2. ¬s → ((p ∧ r) → u)
3. u → (s ∧ ¬t)
Prove from the previous arguments. p → q

Hey guys, I am really lost, so far I have a few arguments but not sure if they're correct. Can you please give some arguments to get this solved or a game plan using Inference laws and equivalences.


From 1. by de Morgan $(r\lor q)\land (\neg a\lor \neg s)$, hence $\neg s$ (and $r\lor q$ or equivalently $(\neg r)\to q$ will be needed later). Especially, $s\land \neg t$ is false hence from 3. by contraposition $\neg u$ Then from 2. by modus ponens $(p\land r)\to \neg u$ and by contraposition $\neg(p\land r)$, i.e. $\neg p\lor \neg r$ or equivalently $p\to\neg r$. Together with $(\neg r)\to q$, we get $p\to q$.

  • $\begingroup$ Thanks friend! Can you explain 1. and is that an typo? $\endgroup$ – DrJonesYu Feb 14 '14 at 14:05

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