Equivalence of universal closure of formulas and implication 
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...
 A: As my previous answer was wrong, this will instead be an explanation to why it was wrong and what was the correct answer. 

I assumed that the truth value of the individual formulas of
  $A^\forall$ must be the same as those of $B^\forall$ since $A^\forall$
  is satisfied if and only if $B^\forall$ is satisfied.

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. 

either $A→B$ or $B→A$ is true $\tag{5}$

