Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x))]$ Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x)]$
I cannot think if there is a difference between the first implication and the last one. Therefore, I think the proposition is true, although I got someone saying that the proposition is false, again, for the proposition to be false, there must be a difference between $(∀x)A(x) → (∀x)B(x)$ and$(∀x)(A(x) → B(x))$ and I just can't think in one.



This is one of the solution that I tried, but now I think that this is wrong, because if $(∀x)A(x)$ is false in the first implication, then so It's in the last one.
 A: 
I cannot think if there is a difference between the first implication and the last one.

There is a difference.  The first proposition says 'If everything is an $A$, then everything is a $B$'. The second proposition says that 'everything that is an $A$ is a $B$' ... which we typically phrase as 'All $A$'s are $B$'s'.
Now, if everything in the domain is an $A$, then by the first proposition everything would also be a $B$, and hence everything would be both an $A$ and a $B$, and it would therefore also be true, as stated by the second proposition, that every $A$ is a $B$. So yes, oif everything is an $A$, then both propositions are true, and the whole statement that is given to you would be true as well.
However, what if only some of the things in the domain are an $A$?  Then the first proposition does not 'apply'... that is, it would not be true that everything is an $A$, and hence you cannot conclude that everything is a $B$ .. let alone that anything is a $B$ at all.
Maybe somewhat counterintuitively, in this case (where some things are an $A$ but others things are not an $A$), the first proposition would still be true: whenever the antecedent ('if' part) of a conditional is false, then the whole conditional is true (indeed, since the 'if' part does not 'apply', the proposition is not making any claims about everything being a $B$, and hence the proposition cannot be said to make a false statement).
However, if we also assume that there are in fact no $B$'s at all in the world, this would also mean that the second proposition is false, since those things that are an $A$ would not be a $B$.
And so, with a world where some things are $A$, but not everything is an $A$, but where there are no $B$'s at all, the first proposition is true, but the second one is false. Hence, the whole statement would be false.
This tells us that the statement given to you can be true or false depending on the world/domain in which you evaluate it. Indeed, when the title of your question asks whether the statement is true or false, we just can't tell.  However, asking whether some logical statement is true or false is in fact a rather unusual question, and I wonder whether this was really the question you had to answer. More typically, we would ask: is some statement logically true (or: is it necessarily true), meaning: is it true no matter what the domain is like? (roughly: 'is it always true?')  And if that is the question (I bet it is), then the answer is no ... because as I showed above, we can find a world (sometimes called a counterexample) where the statement is false. So, it is not a necessary or logical truth.
A: I think, the proposition is false.
Let $S := \{1,2,3,4,5\}$ and $f : S \to \mathbb{R}$ be a function. If $\forall x \in S: f(x) > 0$, than surely $\forall x \in S : \sum_{i=1}^{x} f(i) > 0$. In this case, $\forall x \in S: f(x) > 0 \Rightarrow \sum_{i=1}^{x} f(i) > 0$ does not necessarily hold: If $f(2) = 1$ and $f(1) = -2$, than $f(2) > 0$, but $\sum_{i=1}^{2} f(i) < 0$.
A: Are you familiar with the semantic tableaux method? that way it is easy to find if there's a couterexample.
Consider an element of the domain $a$ that is not $A$, so $\lnot Aa$, what would then be the truth value of the formula $\forall x Ax \to \forall x Bx$?
Now take another element $c$ such that: $Ac$ and $\lnot Bc$.
What does that tell us about the truth value of the consequent of your formula?
