Proof by contradiction using counterexample Why can't we use one counterexample as the contradiction to the contradicting statement?
Example:
Let a statement be A where a-->b.
We can prove A is not true by finding a counter example.
Now, in another space and time, Let a new statement be B where it is the same as a-->not b.
Why can't we prove B is not true by finding a counter example?
Thus, a contradiction and A is true?
 A: Instead of having both $A$, $B$, $a$ and $b$, let's make some of them $p$ and $q$.
Also, since you're talking about counterexamples, there must be a quantifier somewhere, so I guess you actually mean something like
$$ A \equiv \forall x\,(p(x)\to q(x)) $$
$$ B \equiv \forall x\,(p(x)\to \neg q(x)) $$
Having a counterexample to $A$ means that we have a particular $x_0$ such that $p(x_0)$ is true but $q(x_0)$ is false.
Similarly, having a counterexample to $B$ means that we have an $x_1$ such that $p(x_1)$ is true and $q(x_1)$ is also true.
There's no contradiction in having both counterexamples at the same time, because $x_0$ and $x_1$ will in general be different things -- as evidenced by the fact that $q$ is false for one of them and true for the other.
And also,just knowing these two counterexamples doesn't tell us anything about whether $p(x)$ is true for any other $x$, different from $x_0$ and $x_1$. There may be lots of other $x$s that $p$ is false for. Or there might not -- the example doesn't tell us anything either way.
A: You can disprove the statement $a \implies \sim b$ using a counterexample if you can find one.  However, this does not prove the statement $a \implies b$.  This is because the negation of $a \implies \sim b$ isn't $a \implies b$.
Recall that the negation of an implication is not an implication.  $\sim (a \implies b) \equiv a \wedge \sim b$.  In words, this says that the negation of "a implies b" is equivalent to "a and not b".
You are right that if a statement $B$ is the negation of a statement $A$, then $A$ being false implies $B$ is true, and $B$ being false implies $A$ is true.  The problem is, the negation of an implication is not an implication.
A: Remember that:
$(a\implies b) \equiv (\neg a \lor b)$
Hence: $\therefore \neg (a\implies b) \equiv (a \land \neg b)$
The counter example of the implication is an event where $a$ happens and not $b$.
${\begin{array}{|c|c|c|c|}\hline a & b & a\to b & a \land \neg b \\ \hline\hline F & F & T & F \\\hline F & T & T & F \\ \hline T & F & F & \color{red}{T} \\\hline T & T & T & F \\ \hline  \end{array}}$
