A logic problem: $[\lnot q \to (r \lor w), (\lnot q \land p), (p \to \lnot w)] \to r$ 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

 A: 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}
A: This seems rather easy:

*

*$\lnot q \land p$ (hypothesis 2).

*$\lnot q$ (from 1)

*$\lnot q \to (r \lor w)$ (hypothesis 1).

*$r \lor w$ (modus ponens, from 2 and 3).

*$p$ (from 1)

*$p \to \lnot w$ (hypothesis 3)

*$\lnot w$ (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).
