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Why is the contrapostive of an implication equivalent to its normal truth table? i.e. why is this the case:

$$ \begin{array}{c|l|c} p & q & \text~p \implies \text~q \\ \hline 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} $$

Given that the nomal implication table is:

$$ \begin{array}{c|l|c} B & A & B \implies A \\ \hline 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} $$

Specifically, in the first table and first row: p = 1, thus ~p = 0; q = 0, thus ~q = 1. Given these, if one enters these values ( B=0 and A=1) into the second, basic implication table, then the statement is true.

An example would help. I cannot grasp the meaning so I can't really think of any good examples.

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you sure about your resource ? – ashley Dec 11 '12 at 18:58

Your first truth table isn't correct: the first and second rows have the wrong truth value. For it to be true, the heading should be $\sim q \implies \sim p$.

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Ah, Ha! A fool I am. So the counterpositive not only switches the truth of the values, but also flips them. Hence counter-positive.Thanks. – user1711997 Dec 11 '12 at 19:10

Think about the following two statements:

If you are a cat, then you are an animal.


If you are not an animal, you are not a cat.

These statements are contrapositives of each other, but they express the same fact: namely that being an animal is a necessary condition of being a cat. Another compatible way of thinking about it is that "cats" are a subset of "animals".

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