Does $\exists x\in\Bbb{R}$, such that, $\forall y\in\Bbb{R}, x<y \implies e^{f(x)-g(x)}<e^{f(y)-g(y)}$ make sense?

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This is the problem I have:

In this problem, assume all functions have domain $\Bbb{R}$. I will define two new concepts.

  1. We say that $f$ is nicer than $g$ if $$\exists x \in \Bbb{R}, s.t. \forall y \in \Bbb{R}, x < y \implies e^{f(x) - g(x)} < e^{f(y) - g(y)}$$
  2. We say that $f$ is stronger than $g$ if $$\forall x \in \Bbb{R}, \exists y \in \Bbb{R}, s.t. x < y \text{ AND } e^{f(x) - g(x)} < e^{f(y) - g(y)}$$

In sentence 1, it says $\exists x \in \mathbb{R}$, such that, $\forall y \in \Bbb{R}, x<y \implies e^{f(x)-g(x)} < e^{f(y)-g(y)}$. I am confused what this actually means. Does it mean that, for any $x$ that is in $\Bbb{R}$, we need to select those $y$ which are greater than $x$?

If "such that" was placed after $\forall y \in \Bbb{R}$, then this sentence would be: $$\exists x \in \Bbb{R}, \forall y \in \Bbb{R},\text{ such that, }x<y \implies e^{f(x)-g(x)} < e^{f(y)-g(y)}$$ and wouldn't make any sense, since that would imply that every number real number taken as $y$ is greater than arbitrary $x$.

My argument is that, since "such that" is placed right after $\exists x \in \Bbb{R}$, we are only considering those $y$ that are greater than $x$ and hence $x<y$ is true and makes sense.

I am a little confused, so kindly throw some light on if I am correct or wrong, please.