TITLE: Propositional Logic Help: $(\neg p \wedge (p \vee q)) \rightarrow q $ is a tautology
QUESTION [6 upvotes]: I need to prove that $(\neg p \wedge (p \vee q)) \rightarrow q $ is a tautology using Laws of Logic (not truth tables).
This is what I tried: 
$\equiv (( \neg p \wedge p) \vee (\neg p  \wedge q)) \rightarrow q \\
\equiv (F \vee (\neg p  \wedge q)) \rightarrow q \\
\equiv (\neg p  \wedge q) \rightarrow q \\
\equiv (F) \rightarrow  q \\
\equiv T $
Is this logically correct?
The laws I used in order were: Distributive, then negation, and identity. My only issue is with the last step where I know the truth values of $(\neg p  \wedge q)$ are all $F$ but I dont know what law it uses.
Please Help!

REPLY [6 votes]: Hint: We have that
$$
(\neg p \land q)\to q \equiv (p\lor \neg q)\lor q \equiv p \lor (\neg q\lor q). 
$$
You should be able to take it from there.