# Math Help - p and q logic proof

1. ## 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.

2. Using a truth table.

3. (P↔Q) → (Q ↔R)

and

Q → (¬R → ¬P)
These formulas are not equivalent: consider P = Q = False and R = True.

4. 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