∼(p ∨∼q) ∨ (∼p ^ ~ q) ≡ ~p
Please help I don't know where to start.
These are the laws I need to list in each step when simplifying.
Commutative laws: p ∧ q ≡ q ∧ p
p ∨ q ≡ q ∨ p
Associative laws: (p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
Distributive laws: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Identity laws: p ∧ t ≡ p
p ∨ c ≡ p
Negation laws: p ∨ ∼p ≡ t
p ∧ ∼p ≡ c
Double negative law: ∼(∼p) ≡ p
Idempotent laws: p ∧ p ≡ p
p ∨ p ≡ p
Universal bound laws: p ∨ t ≡ t
p ∧ c ≡ c
De Morgan’s laws: ∼(p ∧ q) ≡ ∼p ∨ ∼q
∼(p ∨ q) ≡ ∼p ∧ ∼q
Absorption laws: p ∨ (p ∧ q) ≡ p
p ∧ (p ∨ q) ≡ p
Negations of t and c: ∼t ≡ c
∼c ≡ t
$\endgroup$ 12 Answers
$\begingroup$$\neg(p \vee \neg q) \vee (\neg p \wedge \neg q) \equiv \neg p$
$\begin{align} \neg(p \vee \neg q) \vee (\neg p \wedge \neg q) & \equiv (\neg p \wedge \neg \neg q) \vee (\neg p \wedge \neg q) & \text{D'Morgan} \\ & \equiv (\neg p \wedge q) \vee (\neg p \wedge \neg q) & \text{Double Negation} \\ & \equiv \neg p \wedge (q\vee \neg q) & \text{Distribution} \\ & \equiv \neg p \wedge \top & \text{Conjunctive Negation} \\ & \equiv \neg p & \text{Identity} \end{align}$
$\endgroup$ $\begingroup$First apply De Morgan to : $∼(p ∨∼q)$, followed by Double Negation on $∼∼q$.
Then apply Distributivity : $(∼p ∧ q) ∨ (∼p ∧ ∼q) ≡ p ∧ (q ∨ ∼q)$ followed by Negation law : $q ∨ ∼q ≡ t$.
Finally, simplify $∼p ∧ t$ with Identity law.
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