Prove/Disprove: If $A^{\forall}$ is logically equivalent to $B^\forall$ (that is, any structure $M$ and assignment $\rho$ satisfies $A^\forall$ if and only if it also satisfies $B^\forall$), then either $A\to B$ or $B\to A$ is true (satisfied in any structure/assignment).
My intuition says it's false, as I think $A^\forall$ and $B^\forall$ can be equivalent in that for some assignments $A$ fails and $B$ is true while for others $B$ fails and $A$ is true, which would mean neither $A\to B$ nor $B\to A$ are true, but I'm stuck on this for a long time unable to find a suitable counter example...
As my previous answer was wrong, this will instead be an explanation to why it was wrong and what was the correct answer.
However, consider the case in which a single non quantified formula of $A^\forall$ is false. Now, since $A^\forall$ is false iff $B^\forall$ is false, there must be at least one false formula in $B^\forall$ .
For the case in which there is only one, the truth values of the formula (1) could be that of (3) and the values of (2) those of (4) .
$A\rightarrow B\tag{1}$
$B\rightarrow A\tag{2}$
$F\rightarrow T\tag{3}$
$T\rightarrow F\tag{4}$
In such a case (1) is true while (2) is false. Hence the truth value of the individual formulas of $A^\forall$ need not be the same as those of $B^\forall$ .
The statement (5) however, is still true, since at least one of either $A→B$ or $B→A$ has to be true; because the same $B$'s and $A$'s are used.