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Why is the implication $P \Rightarrow Q$ false if and only if $P$ is true and $Q$ is false ?

Is this because if $P$ implies $Q$, where $P$ is true and $Q$ is false, then $Q$ is also true by definition of the implication ?

The exact definition of the implication $P \Rightarrow Q$: if $P$ then $Q$ is also true ?

So given an implication $P \Rightarrow Q$ where $P$ is true and $Q$ is false it must be false, since no true statement can never imply a false statement, since this would mean the false statement would be true, which it is not ?

Shuzheng
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  • What do you mean by "why"? Do you want a proof of why your statement is true or do you want to understand the reason why the axioms were chosen so that your statement was true? – madprob Jan 12 '14 at 21:46
  • Why? Because this question has been asked like a zillion times before on this site. – Asaf Karagila Jan 12 '14 at 21:47
  • To name a few: http://math.stackexchange.com/q/48161/ and http://math.stackexchange.com/q/53777/ (with its links) http://math.stackexchange.com/q/137890/ and probably several more. – Asaf Karagila Jan 12 '14 at 21:48
  • $p\rightarrow q \equiv \neg p \lor q$. Now write the truth table. –  Jan 12 '14 at 22:26

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