Using the substitution theorem and the important equivalences (handout) show the following equivalence. Use only one substitution/equivalence rule (such as absorption) per step and justify each step by name'

Respuesta :

Answer:

Explanation:

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