Way to prove that two propositional statements are equivalent with and without truth table?

Question

Using the statement $((p \rightarrow q) \wedge p) \rightarrow q$ and $(p \wedge ((( \neg p \vee s) \wedge ( \neg p \vee \neg s)) \vee q)) \rightarrow q$.

I tried doing separately as LHS and RHS and simplified LHS to $( \neg p \wedge p \vee q \wedge p) \rightarrow q$,

however I'm unsure how to eliminate S from RHS.

Answer 1

HINT

Use:

Adjacency

$(p \lor q) \land (p \lor \neg q) \equiv p$

If you don't have adjacency:

$(p \lor q ) \land (p \lor \neg q) \equiv$

$p \lor (q \land \neg q) \equiv$

$p \lor \top \equiv$

$p$