Material implication (rule of inference)

Suppose we are given that P \to Q. Then we have \neg P \lor P by the law of excluded middle (i.e. either P must be true, or P must not be true). Subsequently, since P \to Q, P can be replaced by Q in the statement, and thus it follows that \neg P \lor Q (i.e. either Q must be true, or P must not be true). Suppose, conversely, we are given \neg P \lor Q. Then if P is true, that rules out the first disjunct, so we have Q. In short, P \to Q. This can also be expressed with a truth table: == Example ==

An example: we are given the conditional fact that if it is a bear, then it can swim. Then, all 4 possibilities in the truth table are compared to that fact. • If it is a bear, then it can swim — T • If it is a bear, then it can not swim — F • If it is not a bear, then it can swim — T because it doesn’t contradict our initial fact. • If it is not a bear, then it can not swim — T (as above) Thus, the conditional fact can be converted to \neg P \vee Q, which is "it is not a bear" or "it can swim", where P is the statement "it is a bear" and Q is the statement "it can swim". ==The equivalence does not hold in intuitionistic logic==

Intuitionistic logic does not treat P \to Q as equivalent to \neg P \vee Q because : P \to Q \cancel{\Rightarrow} \neg P \or Q Given P \to Q, one can constructively transform a proof of P into a proof of Q. In particular, P \to P holds in intuitionistic logic. If P \to Q \Rightarrow \neg P \or Q would hold, then \neg P \or P could be derived. However, the latter is the law of excluded middle, which is not accepted by intuitionistic logic (one cannot assume \neg P \or P without knowing which case applies). ==References==