Thread: p and q logic proof

Hello all. I am trying to prove

(P↔Q) → (Q ↔R)


and



Q → (¬R → ¬P)


are logically equivalent, by way of proof. Rules are DeMorgan's Laws may not be used. Any help is appreciated. Thanks.

Using a truth table.

(P↔Q) → (Q ↔R)

and

Q → (¬R → ¬P)

These formulas are not equivalent: consider P = Q = False and R = True.

Originally Posted by cdoil

Hello all. I am trying to prove

(P↔Q) → (Q ↔R)


and



Q → (¬R → ¬P)


are logically equivalent, by way of proof. Rules are DeMorgan's Laws may not be used. Any help is appreciated. Thanks.

[(P↔Q) → (Q ↔R)] logically implies [Q → (¬R → ¬P)] and hence provable.

(meaning, by assuming [(P↔Q) → (Q ↔R)] we can prove [Q → (¬R → ¬P)] )

But [Q → (¬R → ¬P)] does logically implies [(P↔Q) → (Q ↔R)] and thus not provable