I have just started studying implication in mathematics. So, you'll probably have an idea of where my confusion lies right from the get go.
In the truth table, where $A \implies B$ we obtain this result:
A | B |A->B
___________
T | T | T
T | F | F
F | T | T
F | F | T
Now, my confusion here is regarding why $A\implies B$ given A is false, and B true. The others I understand. The first and last one are obvious, the second one implies, to me anyway, that given A implies B, the truth of B rests upon the truth of A, B is false, A is True, which cannot be, thus not B given A is false.
Now... then, why is B true despite the fact that A is false? Or rather, why is the statement B given not A, True. The truth of B is implied from the truth of A, but A is a false statement. So it's very contradictory that this may be, and in fact, if you go through a real life example, you would find that the third statement seems to be false.
Say $A$ is the statement that it is dark outside, and $B$ the statement that the sun is not on our side of the earth. If it is not dark outside, than stating that the sun is on the other side of the planet must be wrong (in my example i'm going to claim that the sun is the only light source).
Side note:
I just had a thought, if A is false, does that mean you can conclude... mmm, whatever? I mean, B is implied via the truth of A, if A is false, B is NOT implied, but then B could be EITHER true or false, I suppose. Is this correct?
However, would that not mean that A does not imply anything when false. Would that not set the 3rd and 4th statement to be undefined?
I just had a thought, if A is false, does that mean you can conclude... mmm, whatever?
Yes. Exactly that. $\endgroup$