TITLE: Can't understand the solution given on logical equivalence
QUESTION [8 upvotes]: I'm working on this question:
Use Theorem 2.1.1 to verify the logical equivalence  $∼(∼p ∧ q) ∧ (p ∨ q) ≡ p.$
I'm guessing that I either have a flawed understanding either about the distributive law or absorption law or both.
These were my steps and was stuck:
$∼(∼p ∧ q) ∧ (p ∨ q) ≡ (∼(∼p) ∨ ∼q) ∧ (p ∨ q)$ <--By De Morgan’s laws 
$≡ (p ∨ ∼q) ∧ (p ∨ q)$ <--by the double negative law
$≡{ ( (p ∨ ∼q) ∧ p ) ∨ ( (p ∨ ∼q) ∧ q ) }$ <--By distributive law
$≡{ p ∨ ( (p ∨ ∼q) ∧ q }$ <--By absorption law
$≡{ p ∨ ( (q ∧ p) ∨ (q ∧ ∼q) ) }$ <--By distributive law
$≡{ p ∨ ( (q ∧ p) ∨ 0 ) }$ <--By negation law
$≡{ p ∨ (q ∧ p) }$ <--By identity law
$=p$<--By adsorption law (Edited after a comment, thus I have edited the question)
The answer was:
$∼(∼p ∧ q) ∧ (p ∨ q) ≡ (∼(∼p) ∨ ∼q) ∧ (p ∨ q)$ by De Morgan’s laws
$≡ (p ∨ ∼q) ∧ (p ∨ q)$ by the double negative law
$≡ p ∨ (∼q ∧ q)$ by the distributive law
$≡ p ∨ (q ∧ ∼q)$ by the commutative law for ∧
$≡ p ∨ c$ by the negation law
$≡ p$ by the identity law
So my question how did the solution jump from $(p ∨ ∼q) ∧ (p ∨ q)$  to  $p ∨ (∼q ∧ q)$ just using distributive law?

REPLY [13 votes]: Distributive laws states that
$${\displaystyle p\vee (q\wedge r)\equiv (p\vee q)\wedge (p\vee r)}\tag{1}$$
$${\displaystyle p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)}\tag{2}$$
Here we are applying $(1)$ that $(\color{red}p\vee \color{blue}q)\wedge (\color{red}p\vee \color{green}r)$ implies $\color{red}p\vee (\color{blue}q\wedge \color{green}r)$.
That $(\color{red}p∨\color{blue}{∼q})∧(\color{red}p∨\color{green}q)$ implies $\color{red}p∨(\color{blue}{∼q}∧\color{green}q)$ which is distributing $p$ out.