Difference between $(\forall x A(x)) \implies (\forall x M(x))$ and $\forall x (A(x) \implies M(x))$ I don't understand the difference between
$$(\forall x A(x)) \implies (\forall x M(x))$$
and
$$\forall x (A(x) \implies M(x))$$ 
How do the additional quantifiers change the statement?
 A: In the second statement, you're relating each specific $A(x)$ to the corresponding $M(x)$, while in the first statement you're relating the statement $A$ as a whole to the statement $M$ as a whole.
Example: let's say $\forall$ ranges over the natural numbers, $A(x)$ means "$x$ is even" and $M(x)$ means "$x$ is prime". Then $\forall x\, A(x)$ and $\forall x\,M(x)$ are both false, so the first statement is true. However, $A(4)\implies M(4)$ is false, so the second statement is false.
A: Let $A(x)$ be that $x$ likes the movie. Let $M(x)$ be that $x$ is happy.
The first proposition states that everyone is happy if everyone likes the movie. The second proposition states that each person is happy if he likes the movie.
A: Another example, suppose we are restricted to natural numbers $0, 1, 2, \dots$ , and suppose there is some unspecified natural number $k$.
Let $A(x) \equiv (x \ge k)$
Let $M(x) \equiv (x \times k = 0)$
$\forall x Ax \to \forall x M x$ is true because $\forall x Ax$ implies that $k=0$.
However $\forall x (Ax \to Mx)$ is not provable because, $x = 1, k = 1$ is a counterexample.
If you are up to it, it could be noted that $\forall x Ax \to \forall x Mx$ is equivalent to $\forall x ((\forall y Ay) \to Mx)$, so you are also comparing
$$\forall x (Ax \to Mx)$$
$$\forall x ((\forall y Ay) \to Mx)$$
Since $(\forall y Ay)$ is a stronger claim than $Ax$, the first statement is a stronger statement than the second; that is, the second statement is implied by the first statement.  So the only way to find a counter example is to find a situation where the first statement is false but the second is true.
So in order for the first statement to be false, there has to be some case where $Mx$ is false and $Ax$ is true.  $Mx \equiv (x \ge 11)$ and $Ax \equiv (x \ge 10)$ is such a situation, when $x=10$ then $A$ holds and $M$ fails.  Then for the second statement to hold, all you need is $\forall y Ay$ to be false.
