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Pleae forgive the very basic question, but I know nothing really of formal logic and so would appreciate some feedback.

The truth table defining the implication operator

P    Q    P implies Q
T    T         T 
T    F         F 
F    T         T 
F    F         T 

together with the negation operator ~ defined in the obvious way enables one to construct the following table for ~Q $\implies$ ~P:

P  Q  ~P  ~Q  ~Q implies ~P
T  T   F   F        T 
T  F   F   T        F 
F  T   T   F        T 
F  F   T   T        T 

Evidently, truth values for ~Q $\implies$ ~P are the same as those for P $\implies$ Q. Is this enough to prove that $P \implies Q$ if and only if ~Q $\implies$ ~P ? My thinking is that yes, it is, because I belive that the logical operators involved are defined by their respective truth tables and this being the case the observations above should be sufficient to prove the equivalence.

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up vote 2 down vote accepted

Yes what you're saying is indeed the case and is the idea behind the technique of proof by contrapositive.

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In the context of formal logic, such will usually not consist of a proof... or more perhaps more clearly, a formal proof. It will "show" that iff (P ⟹ Q), then (~Q ⟹ ~P) comes as valid. For informal logic settings, that will usually suffice, and it could suffice as an argument of some sort in a formal logic setting given that you've already established the completeness metatheorem of classical propositional logic. However, it does not consist of a formal proof, since in formal logic, a formal proof gets defined something like "a sequence of well-formed formulas (often just called "formulas") such that every well-formed formula is either an axiom, a hypothesis made under a certain scope which will get discharged eventually, or a well-formed formula permissible according to the rules of inference and well-formed formulas already in the proof." What you've given above, does not give us such a sequence of well-formed formulas.

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