Just fill in the truth table like this:
P | Q | P → Q | ~Q → ~P | (P → Q) ↔ (~Q → ~P)
––|–––|–––––––|––––––––––|––––––––––––––––––––
T | T | T | T | T
T | F | F | F | T
F | T | T | T | T
F | F | T | T | T
As you see, these two have identical truth values, and they are _logically equivalent_. It means that `(P → Q) ↔ (~Q → ~P)` is a tautology, not `~Q → ~P`. If two things are logically equivalent, they create a tautology when in the form of a biconditional.
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