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Will somebody please solve this completely for me? $$[\lnot q \to (r \lor w), (\lnot q \land p), (p \to \lnot w)] \to r$$

I did it but got stuck at the end

Here is my incomplete answer enter image description here

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  • $\begingroup$ I think you may be overthinking this. You have $\neg q$, so you have $r \vee w$; you have $p$, so you have $\neg w$; you have $r \vee w$, so you have $\neg w \to r$; so at last you have $r$. $\endgroup$ – Brian Tung Jun 28 at 8:03
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Hint

Based on what you had done in the attempt, I believe you want to prove this use logical equivalence, the ',' should means 'and'.

Until you write the following, everything else is correct

This is just the wrong way to apply the de morgan's law, apply it correcly we should get $$\boxed{[(\neg q\land\neg r\land\neg w)\lor (q\lor \neg p)\lor(p\land w)]\lor r}$$ Try again see if you can prove this is a tautology.

Answer

\begin{align} &[(¬q→(r∨w))\land(¬q∧p)\land(p→¬w)]\to r\\ \equiv&[(q\lor r\lor w)\land(¬q\land p)\land(\neg p\lor \neg w)]\to r\\ \equiv&\boxed{(\neg q\land \neg r\land \neg w)\lor q\lor \neg p\lor(p\land w)\lor r}\\ \equiv&(\neg q\land \neg r\land \neg w)\lor q\lor ((\neg p\lor p)\land(\neg p\lor w))\lor r\\ \equiv&(\neg q\land \neg r\land \neg w)\lor q\lor (\top\land(\neg p\lor w))\lor r\\ \equiv&(\neg q\land \neg r\land \neg w)\lor q\lor \neg p\lor w\lor r\\ \equiv&((\neg q\lor q)\land (\neg r\lor q)\land (\neg w\lor q))\lor \neg p\lor w\lor r\\ \equiv&(\top\land (\neg r\lor q)\land (\neg w\lor q))\lor \neg p\lor w\lor r\\ \equiv&((\neg r\lor q)\land(\neg w\lor q))\lor \neg p\lor w\lor r\\ \equiv&((\neg r\lor q\lor w)\land(\neg w\lor q\lor w))\lor \neg p\lor r\\ \equiv&((\neg r\lor q\lor w)\land\top)\lor \neg p\lor r\\ \equiv&\neg r\lor q\lor w\lor \neg p\lor r\\ \equiv&q\lor w\lor \neg p\lor \top\\ \equiv&\top \end{align}

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This seems rather easy:

  1. $\lnot q \land p$ (hypothesis 2).
  2. $\lnot q$ (from 1)
  3. $\lnot q \to (r \lor w)$ (hypothesis 1).
  4. $r \lor w$ (modus ponens, from 2 and 3).
  5. $p$ (from 1)
  6. $p \to \lnot w$ (hypothesis 3)
  7. $\lnot w$ (modus ponens from 5,6)

and then deduce $r$ from $4$ and $7$. (Which requires the law of the excluded middle; systems will vary in how to deduce it).

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