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.
Best Answer
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 $\sim q \implies \sim p$.