Determine whether the predicate $\forall x \bigl(P(x) \leftrightarrow Q(x)\bigr)$ is logically equivalent to the predicate $\forall x P(x) \leftrightarrow \forall x Q(x)$.
I would be willing to wager that the statement is false, and I would like a counterexample.
Here is the reason that I think the statement is false.
If $\forall x \bigl(P(x) \leftrightarrow Q(x)\bigr)$, for every $x_{\circ}$ in the domain of discourse, $P(x_{\circ}) \leftrightarrow Q(x_{\circ})$ is a true statement. Now, assume that $\forall x P(x)$. If there were one element $y$ in the domain of discourse such that $\neg{Q(y)}$ is true, $\forall x P(x) \leftrightarrow \forall x Q(x)$ would be false.
Best Answer
Counterexample: Let the domain of discussion be $\mathbb{N}$. Let $P(x) = x$ is even. Let $Q(x) = x$ is odd.
In this case $\forall x \bigl(P(x) \leftrightarrow Q(x)\bigr)$ will be false, and $\forall x P(x) \leftrightarrow \forall x Q(x)$ will be true.
EDIT: We can show that "Every natural number is even if and only it is odd" is false. And that "Every natural number is even if and only if every natural number is odd" is true.
In this case, $\forall x \bigl(P(x) \leftrightarrow Q(x)\bigr)$ will be false since $P(x)$ and $Q(x)$ will always differ.
EDIT: A natural number cannot be both even and odd.
Note that if both $A$ and $B$ are false, then $A\leftrightarrow B$ is true. In this case, both $\forall x P(x)$ and $\forall x Q(x)$ are false. Therefore, the biconditional $\forall x P(x) \leftrightarrow \forall x Q(x)$ must be true.
EDIT: "Every natural number is even" is false. As is "Every natural number is odd." Therefore, "Every natural number is even if and only if every natural number is odd" is true.
Aside: In general (for any P and Q), we can show that $\forall x \bigl(P(x) \leftrightarrow Q(x)\bigr) \to \forall x P(x) \leftrightarrow \forall x Q(x)$.