For example, if there is theorem that says: "$[x] = [y] \iff x \sim y$," and I am asked to prove $[(a,b)] = [(c,d)]$
Is it enough to show that $[(a,b)] = [(c,d)] \implies (a,b) \sim (c,d)$, because of the theorem that says "$[x] = [y] \iff x \sim y$?"
Is it valid to assume $[(a,b)] = [(c,d)]$ and prove $(a,b) \sim (c,d)$, even though $[(a,b)] = [(c,d)]$ is the thing we are trying to prove in the first place?
No, it isn't OK. For instance, suppose you are considering a 2D geometric shape and $A$ is the statement "The shape is a square" and $B$ is the statement "The shape has 4 equal sides and 4 right angles." It is true that $A \iff B$.
However, if you prove that $A \implies B$, that doesn't mean you've proven $A$. All you've shown is that if your shape is a square, then it has 4 equal sides and 4 right angles.
To prove $A$ using your theorem, you want to show that $B$ is true. In the shape example, if you were trying to prove your object was a square, you would prove that your object had 4 equal sides and 4 right angles.
In your example, you want to show $(a,b)$~$(c,d)$. Then apply the theorem that gives you [(a,b)]=[(c,d)].
