In classical logic, $B\rightarrow A$ is equivalent to $\lnot A \rightarrow \lnot B$, so you will simply use the simplest definition of $A\leftrightarrow B$, which is to use $B\rightarrow A$. In intuitionistic logic, the contraposition is not generally derivable. You have $(B\rightarrow A)\rightarrow(\lnot A\rightarrow \lnot B)$, but not $(\lnot A\rightarrow \lnot B)\rightarrow(B\rightarrow A)$. Hence, your definition of equivalence is weaker - and doesnt really make sense, it only makes sense if you assume $A\vee \lnot A$.
And if you look at it computationally, $A\leftrightarrow B$ means you have terms $u^{A\rightarrow B}$, $v^{B\rightarrow A}$, such that if you have an object $a^A$ you always get an object $b^B$ and vice versa. With your definition, you would have $u^{A\rightarrow B}$, but $v^{(A\rightarrow\bot)\rightarrow(B\rightarrow\bot)}$ where $\lnot A = A\rightarrow\bot$ and $\bot$ can be defined as $0=1$ (or any other type of which no term is derivable), hence, $v$ maps terms $t^{A\rightarrow\bot}$ that map terms of type $A$ on something not derivable onto terms $u^{B\rightarrow\bot}$ mapping terms of type $B$ on something not derivable.
This isnt what we usually want from equivalence.