Question about demonstration of the type $A\Leftrightarrow B$

Question

I have an important question when it is asked to us to proove affirmation/theorem wich look likes this $A\Leftrightarrow B$
I know that in order to proove $A\Leftrightarrow B$ i must proove first that: $A\Rightarrow B $ and after i must proove $A \Leftarrow B $

Let suppose i ve succeeded in prooving $A\Rightarrow B$ (1), so it lets me to proove: $B \Rightarrow A $.
We know that to proove $B \Rightarrow A $ is equivalent to proove the contraposition: $no(A)\Rightarrow no(B)$
So now let suppose that i choose to proove $no(A)\Rightarrow no(B)$ by absurd and that my absurd assumption looks like this: $A\Rightarrow no(B)$.
But $A\Rightarrow no(B)$ is in contradiction to (1) that i've allready prooved. So now my demonstration is finish.

EDIT1: Let suppose too that i ve succeeded to prove that $ A⇒no(B)$ isn't possible.

Is this correct? Can you give me exemple or counter exemple (simple)?

Thank you.

Answer 1

The negation of $(A \Rightarrow B)$ is $(A \land \neg B).$

If you have already proved $(A \Rightarrow B)$ and you also succeed in proving $(A \Rightarrow \neg B)$, then you have proved that $A$ is false, which means that $A$ actually implies every other statement. In this case $(B \Rightarrow A$) is true if and only if $B$ is false.

Answer 2

There are two flaws in your logic.

First, $A\implies no(B)$ does not contradict $A\implies B$. Both can be true if $A$ never is.

Second, $A\implies no(B)$ is not the negation of $A\implies B$, so your reasoning by absurd assumption is not valid.