How to prove that $(p \land (\neg(\neg p \lor q))) \lor (p \land q) \equiv p$?

Question

My attempt at this mathematical demonstration was as follows:

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) ≡ p ≡ (p ∧ (p ∨ q) ∨ (p ∧ q) ≡ ((p ∨ p) ∧ q) ∨ (p ∧ q) ≡ ((p ∨ p) ∧ q) ∨ (p ∧ q) ≡ (p ∧ q) ∨ (p ∧ q) ≡ p

Answer 1

Via implies form, we have $$\neg\left ( \neg p\lor q \right )\quad\equiv\quad\neg\left ( p\Rightarrow q \right )$$

Answer 2

Via Boolean algebra,

$$p \, \overline{(\bar{p} + q)} + p q = p \bar{q} + p q = p \underbrace{(\bar{q} + q)}_{=1} = \color{blue}{p}$$

Answer 3

If you're asking for a check of what you've done, you've made a mistake already in the first step. We have (from the De Morgan's law) $$\lnot(\lnot p \lor q) \equiv (\lnot\lnot p) \land (\lnot q) \equiv p \land (\lnot q) $$ so your expression after the first transformation should be $$ (p \land (p \land (\lnot q))) \lor (p \land q) $$ See if you can take it from here.

Answer 4

As you know from De Morgan's laws, $\left(p\land \lnot(\lnot p\lor q)\right)\lor(p \land q)\equiv \\ (p\land (p\land\lnot q))\lor(p\land q)\equiv \text{De Morgan's Law}\\ ((p\land p)\land \lnot q)\lor(p\land q)\equiv \text{Associativity}\\ (p\land \lnot q)\lor (p\land q)\equiv \\ p\land (\lnot q\lor q)\equiv \text{ Un-Distributivity}\\ p\land T\equiv \\ p$