Truth value of proposition I have the following propositions, is one of them true?

$\forall_y\in\mathbb{Z}, \forall_x\in\mathbb{Z} (\forall_z\in\mathbb{Z}(x*z=y*z \implies x=y))$
$\forall_y\in\mathbb{Z}, \forall_x\in\mathbb{Z} (\exists_z\in\mathbb{Z}(x*z=y*z \implies x=y))$

It's very confusing, my thinking is that the second one is true but it depends on how you interpret the expressions...
Thank you.
 A: After a mis-step, I think I now read both propositions correctly: as discussed in the comments above by @Max and @MathStudent, the first proposition is false while the second proposition is true.
The first proposition fails because we can choose infinitely many triples $(x, y, z)$ of integers such that $x * z = y* z$ and yet $x \neq y$. Specifically, 
whenever $z = 0$ but $x$ and $y$ have already been chosen to be distinct, then the antecedent is true but the consequent is false. 
To see that the second proposition is true, it might be clearer to consider the two cases for possible choices of $x$ and $y$:


*

*$x = y$

*$x \neq y$


We now need to check there exists a $z \in \mathbb{Z}$ such that $x*z=y*z \implies x = y$.
If (1) holds, then choosing $z = 0$ means that both $x*z=y*z$ and $x = y$ are true, so the whole proposition is true.
If (2) holds, then choosing any $z \in \mathbb{Z} - \{0\}$ will mean that the antecedent $x*z = y*z$ is false and the consequent $x = y$ is false. Since "false implies false", the whole proposition is again true.
