Logic - Quantifiers Why is this proposition always true?

$$\forall x\,\forall y\,\exists z\big((x<z)\to (x\ge y)\big)$$

And where`s the flaw in my thinking: 
You can always choose a $z$ larger than $x$. So the problem can be reduced to: 

For all $x$ and all $y$, it is such that $x\ge y$, 

which obviously isn´t true...


Why is this proposition always true?
$\forall x\,\forall y\,\exists z\big((x<z)\implies (x\ge y)\big)$

Suppose we have $x$, $y$ in $\mathbb{R}$
By the irreflexive property of $\lt$, we have $\neg x\lt x$
Introducing $\lor$, we have $\neg x\lt x \lor x \ge y$
Applying the definition of $\implies $, we have $\neg\neg x\lt x \implies x\ge y$
Removing the double negation, we have $x\lt x \implies x\ge y$
Introducing $\exists$, we have $\exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$
Generalizing, we have $\forall x,y \in \mathbb{R}: \exists z\in \mathbb{R}:[x\lt z \implies x\ge y]$
