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?