Rules of Replacement IV

Question 1

Negation of an Equivalence

Answer 1

~ (p <-> q) <=> (p <-> ~q)
~ (p <-> q) <=> (~p <-> q)

Question 2

Proof by Cases

Answer 2

((p v q) -> r) <=> (p -> r) ^ (q -> r)
((p ^ q) -> r) <=> (p -> r) v (q -> r)

Question 3

Pseudo-Distribution of an Implication

Answer 3

(p -> (q v r) <=> (p -> q) v (p -> r)
(p -> (q ^ r) <=> (p -> q) ^ (p -> r)

Question 4

Reductio ad absurdum

Answer 4

(p -> q) <=> ((p ^ ~q) -> F)