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It assumes that '' P implies Q '' is true and Q is false. How is P? I think that P is false.

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    What happens if P is true? ā€“ 2017-01-28
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    Yep, prove it by contradiction: Q is false, let's assume that P were true. But if P is true then Q is true, contradiction ā€“ 2017-01-28
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    If P is true, then "P implies Q" is false! Contradiction!! ā€“ 2017-01-28

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Yes as you often meet a non-sports person saying that,"Had (If) I played (play) tennis then I am a Wimbledon champion."

Irrespective of this combination of two false clubbed statements, you forgive them for their innocence.

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Truth table of implies $\Rightarrow$

$$\begin{array}{ | c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \text F & \text T & \text T \\ \text F & \text F & \text T \end{array}$$

This shows when P is false implication is true. Also in your case Q is false but implication not. So we can't choose second case.

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If the conditional is True, and the consequent is False, you have the only option that the antecedent is False. p q pā†’q

V V V

V F F

F V V

F F V

Notice that with the conditional True and the consequent (Q) False, you have in the fourth line that the antecedent (P) is False.

I hope I've helped.