Is it true that $((A\rightarrow B)\land(¬A\rightarrow ¬B))↔((¬A) \;\;\text{⊕}\;\; B)$? Is $((A\rightarrow B)\land(¬A\rightarrow ¬B))↔((¬A) \;\;\text{⊕}\;\; B)$ true? I found it's true but I don't know what to use it for besides refactoring. How interesting is the statement A→B if not also ¬A→¬B? For example radiation → energy and not ¬radiation→¬energy is not very interesting is it? The interesting statement would be radiation → energy ^ ¬radiation→¬energy or goal → score ^ ¬goal→¬score so that there is determinism and no other way (you can't score if you don't make a goal). It's not legal if you can score without actually making a goal. 
A→B is not very interesting when B all the time can happen regardless of the truth value of A in any case. 
"Either A or B" is that B can't happen if not A has not happened and that excludes $ A\rightarrow B.$
I have a simple model where I think the bottom is more like YES ⊕ NO (and it doesn't at all look like YES OR NO even though the picture doesn't say that YES and NO can't both be true for the same input(?)

 A: Yes, what you propose is an equivalence: see (1) and (2) to follow...
(1) Note that $$(A \rightarrow B)\land (\lnot A \rightarrow \lnot B)$$
$$\equiv (A\rightarrow B) \land (B \rightarrow A)\tag{contrapositive}$$
$$\equiv A \leftrightarrow B) $$
$$\not\equiv (A\rightarrow B)$$
That is, what you are proposing is a biconditional: $A\;$ if and only if $\;B$.
In a strict implication $\;A \rightarrow B,\;$ $A$ is a sufficient condition for $B$, whereas $\,B$ is a necessary condition for $\,A\,$.
When you want to state a relationship in which condition $A$ is a necessary and sufficient condition for $\,B\,$, then you write $\,A \iff B$ or $A\,\leftrightarrow B$.
But there are many relationships in which one variable implies the other, but not the other way around. If $A \subseteq B$, then we know that $a\in A \rightarrow b\in B$, but we cannot say that it is also the case that $b\in B \rightarrow b \in A$.
(2) Note also that $$\lnot A\; \text{ XOR}\;\; B \equiv (\lnot A \lor B)\land \lnot(\lnot A \land B)\quad\quad\quad\quad\tag{1: unpack XOR}$$
$$\equiv (A \rightarrow B) \land (A \lor \lnot B)\quad\quad\quad\quad\quad\tag{2: $\lnot p \lor q \equiv p \rightarrow q$; 3: DeMorgan's }$$
$$\equiv (A\rightarrow B) \land (\lnot A \rightarrow \lnot B)\quad\quad\quad\tag{4: same reason as 2}$$
$$\equiv (A\rightarrow B) \land (B\rightarrow A) \tag{as shown above}$$
$$\equiv A \leftrightarrow B\quad\quad\quad\quad\quad\tag{see top}$$
So both the left side and the right side of your equivalence (which is an equivalence) are equivalent, as well, to $A\leftrightarrow B$
A: As amWhy showed in his answer left hand side is a biconditional. Let P denote the negation of A. Then, on the right hand side you have P XOR B which is true if and only if P and B have mutually exclusive values. Since P is the negation of A, P and A has mutually exclusive values. Therefore, A and B has the same value in other words A iff B. So, both sides are equal. 
