logical equivalence question about contingency

Question

I want to know if these are two acceptable ways of reaching the same answer or is the first incorrect? Thanks

$$p \rightarrow (p \vee q) \equiv \neg p \vee (p \vee q) \equiv (\neg p \vee p) \vee q $$ $$\equiv \text{true} \wedge q \equiv q = \text{contingency} $$

$$p \rightarrow (p \vee q) \equiv \neg p \vee (p \vee q) \equiv (\neg p \vee p) \wedge (\neg p \vee q) $$ $$\equiv \text{true} \wedge p \rightarrow q \equiv p \rightarrow q = \text{contingency}$$

Answer 1

The first is correct, the second is not.   Association is the right rule to use; distribution is not applicable--distribution does not work like that.

Answer 2

There is no distributive law of the form

$$p \vee (q \vee r) \equiv (p \vee q) \vee (p \vee r)$$

If both your derivations were true, we would have that $p \rightarrow q \equiv q$, which is clearly not the case.