I have a question regarding proof by contrapositive.
I want to show that $A \implies B$.
Proof by contrapositive would work like $\neg B \implies \neg A$.
Is it enough to show that this holds for one specific example? In other words, I dont have to show it for the whole set of possible parameters, right? If I´d have to make a case distinction, would it be a problem to assume different disjunct set of parameters for each case?
$\quad$
To illustrate what I mean:
Imagine I assume $f(x) = c$ minimizes something (means its optimal), if it has a certain value $\alpha$.
$f(x) = c$ is optimal$\implies f(x) = c =\alpha$.
Now, proof my contrapositve would look like this:
$f(x) = c \neq \alpha\implies f(x) = c$ is not optimal
$\quad$
Now, for this proof by contrapositve to work we first assume $f(x) = c \neq \alpha$. We´d have to check to different cases.
$f(x) = c < \alpha$
$f(x) = c > \alpha$
Next thing we would have to show, that this implies $f(x) = c$ is not optimal.
Imagine I choose the first case and assume this minimizes the loss. Now, I can show that if I increase the value $c$ a little bit, the loss gets smaller. That means it was not optimal. But if I decrease the value, the loss gets bigger. But this only holds if $\alpha < 0.5$, f.e.
From my understanding, this shows is sufficient for the first case.
My problem is, If we now go to the second case, we´d basically have to assume the opposite. If we decrease $c$ the loss gets smaller and if we increase $c$ the loss gets bigger but only if $\alpha > 0.5$.
I´m not sure if this is actually allowed to assume $\alpha > 0.5$ for one case and $\alpha < 0.5$.
I dont have to show it for all $\alpha$, right? One case where it doesnt hold its enough? And the fact that we have to assume different $\alpha$ for the case distinction is no problem in itself?