Is every proof an array of implications?

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I want to prove an implication but I don't know how I can do it without using contradiction. Is there a way to rewrite the contradiction in logical terms, only using the implication "$\Rightarrow$" sign?

What I know is that $0\neq c(b)=x(b),\forall b\in B_0$ and $0=x(b),\forall b\in B\backslash B_0$. I also know that $0\neq x_{|A}(b)=x(b),\forall b\in A$ and $x(b)=0,\forall b\in B\backslash A$. I want to show that this implies $A=B_0$. I am not sure if this is necessary to say but $A,B_0$ are subsets of $B$.

To my understanding "constructive" proofs are proofs only using the implication sign.

I was wondering whether the contradiction is just an abbreviation of an array of implications (i.e. proofs by contradictions are also contructive).

To show the equality of the two sets I need to verify the implications under the assumption that $b\in B$

$b\in A\Rightarrow b\in B_0$

$b\in B_0\Rightarrow b\in A$

But I don't know how I can show the first implication constructively.

If $b\in A$ then $x(b)\neq 0$. Then I would have said if I assume $b$ is not in $B_0$ then $x(b)=0$ which is a proof by contradiction.

Is there a way only using implication signs to prove the statement, can the contradiction be written as implication?