Exam Proof Read: $\forall x,~\exists y\ni\forall z,~[(z>y)\implies (z>x+y)]$ I just got my exam back, and the question went like this:

Determine the truth value of the following statement, assuming that $x$, $y$, and $z$ are real numbers. Justify your answer.
  $$\forall x,~\exists y\ni\forall z,~[(z>y)\implies (z>x+y)]$$

So I said, "True. Let $y=(z-x+1)$." I meant "False." I got $\varnothing$ points. :-(
 A: Your counter example is wrong. You can't specify $y$ in terms of $z$ because the existential quantifier which bounds $y$ precedes the universal quantifier which bounds $z$. The bold words are there for you to note that you can't even argue that you could swap the order of the quantifiers.
Since the statement is of the form $\color{green}\forall \exists\forall \text{ condition}$, a counter-example to show it is false is merely an example of a real number that replaces the variable of the green universal quantifier. You then need to prove that the existential statement that arises can't be true.
I suggest the counter example $x=1$. If there was an $y$ that made the statement true, it would also be true for $z=y+1$.
A: Consider the negation of the original statement.
$$ \exists x : \forall y, \exists z : \neg [ (z > y) \implies (z > x+y)],$$
or equivalently,
$$ \exists x : \forall y, \exists z :  [ (z > y) \wedge (z \leq x+y)].$$
Better still, using some algebra, $(z > y) \Leftrightarrow (z+x > x+y)$, the inequalities can be combined:
$$ \exists x : \forall y, \exists z :  [ z \leq x+y < x+z ]. $$
Then by choosing $x=1$, and for any $y$, choose $z = y + 1$, the previous statement is true.  Thus, the original is false (with the same choices 
for $x, y, z$ now providing a counter-example).
Hope this helps!
