Help proving a simple tautology $(p\land(p\to\lnot q))\to(\lnot q\lor p)$ Specifically this  $$(p\land(p\to\lnot q))\to(\lnot q\lor p)$$-
spent a lot of time going down the wrong path

 A: You were almost done, just two more steps are missing in your caluculation to reach tautologicalitly:
$
\vdots\\
\equiv q \lor (\neg q \lor p)\\
\equiv (q \lor \neg q) \lor p \quad\text{(associativity)}\\
\equiv \top \lor p \quad \text{(complement)}\\
\equiv \top \quad \text{(domination)}$
A: The statement is valid. To see this, we can use a proof tree, like so

This was generated here
A: ($p \Rightarrow \sim q $ )
$ \wedge p$
by Modus Ponens
$\Rightarrow   \sim q $
and by adition $\sim q \vee p $
A: Your work until line 6. is correct, but there is no Associative law as $a\lor(b\land c)\equiv(a\lor b)\land c$, so Complement law doesn't apply on line $7$, as when we see $a\lor b\land c$, the logical connective "$\land$" always connect first.

\begin{align}
\unicode{x2714}\hspace{6.3ex}~&\vdots\\
\unicode{x2714}\hspace{2ex}6.~\equiv&\lnot p\lor (p\land q)\lor (\lnot q\lor p)\tag*{Double negation}\\
\unicode{x2718}\hspace{2ex}7.~\equiv&\top\land q\lor(\lnot q\lor p)\tag*{Complement law}
\end{align}

To fix this, we can apply Distributive law instead as following
\begin{align}
&\vdots\\
\equiv&\lnot p\lor (p\land q)\lor (\lnot q\lor p)\tag*{Line $6.$}\\
\equiv&(\lnot p\lor p)\land(\lnot p\lor q)\lor (\lnot q\lor p)\tag*{Distributive law}\\
\equiv&(\top\land(\lnot p\lor q))\lor (\lnot q\lor p)\tag*{Complement law}\\
\equiv&(\lnot p\lor q)\lor (\lnot q\lor p)\tag*{Identity law}\\
\equiv&\lnot p\lor (q\lor \lnot q)\lor p\tag*{Associative law}\\
\equiv&\lnot p\lor \top\lor p\tag*{Complement law}\\
\equiv&\top\tag*{Domination law}
\end{align}
