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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?

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    $\begingroup$ $p \vee (p \wedge q)$ is equivalent to $p$. $\endgroup$ – fish May 17 '20 at 4:26
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    $\begingroup$ Nice work! The latex symbol $\neg$ for negation is \neg. $\endgroup$ – Olivier Roche May 17 '20 at 6:21
  • $\begingroup$ I'll use it next time! Thank you $\endgroup$ – Leon May 18 '20 at 0:53
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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.

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