In other words:
The answer to this question (I have soln), is no, it cannot be, but when I do my testing, I get true ...
Let $X$ represent the set of natural numbers.
Let $p(x)$ - x is $even$
Let $q(x)$ - $x^2$ is $even$
On the left hand side, we have:
For all $x \in \mathbb{N}$, x is even $\to$ $x^2$ is even. This is True.
On the right hand side, we have:
For all $x \in \mathbb{N},$ x is even $\to$ for all $x \in \mathbb{N}$, $x^2$ is even
We know that $x \in \mathbb{N},$ x is even is False, because this is not the case. Numbers alternate odd to even.
Thus, $x \in \mathbb{N},$ $x^2$ is even is also False, since this is not the case.
So on the left hand side, we have $F \to F$ = T.
Overall, we have:
$T \iff T$, which is $T$.
My textbook says it is $F$ however. Where is the mistake?
Try this:
$X=\mathbb{N}$, $p(x)$ is "$x$ is even", $q(x)$ is "$x^2$ is odd."
Then $\forall x\in X, p(x)$ is false. Thus $(\forall x \in X, p(x)) \Rightarrow (\forall x \in X, q(x))$ is true. However, $\forall x \in X, (p(x) \Rightarrow q(x))$ is false because, for instance, $p(2)$ is true but $q(2)$ is false.
The mistake in your reasoning is that you tried to verify the statement using one particular model. But it does not hold in all models, so it is not true.