Prove $[\neg p \land (p \lor q)] \to q$ is TRUE using logic laws

Question

Prove that $$[\neg p \land (p \lor q)] \to q$$ is TRUE using logic laws.

Here is how I solved it:

Starting with
$[\neg p \land (p \lor q)] \to q$
An understood logical equivalence
$[\neg \neg p \land \neg(p \lor q)] \lor q$
Double negation
$[p \land \neg(p \lor q)] \lor q)$
De Morgan's law
$[ p \land (\neg p \land \neg q)] \lor q$
Associative law
$[(p \land \neg p) \land \neg q] \lor q$
Negation law
$(F \land \neg q) \lor q$
Identity law
$F \lor q$
$ q $

The problem is that to my understanding, this doesn't necessarily prove the statement to be true, and I am not sure where I have gone wrong.

Answer 1

You've made a mistake:

$$(\neg p \land (p \lor q))\to q$$ is $$(\neg \neg p \lor \neg(p \lor q)) \lor q.$$

Answer 2

I'd start by distribution, since $\neg p\wedge(p\vee q)$ equivates to $(\neg p\wedge p)\vee (\neg p\wedge q)$, which simplifies through complementation and identity, to $(\neg p\wedge q)$.   (Sometimes accepted as a single step, "complementary absorption".)

Then you can apply implication equivalence, and deMorgan's rule to begin simplifying the whole expression down to the truth constant.