Describe: $\{x\in \mathbb R\mid\forall y\left [(y\in \{t\in \mathbb N\mid t>3\})\to (y>x) \right ] \}$ with $x$ as the only parameter.
The inner part can be rewritten: $(y\in \{t\in \mathbb N\mid t>3\}\to (y>x) \equiv \neg(y\in \{t\in \mathbb N\mid t>3\})\vee (y>x)\overset{?}{\equiv}(y\not\in \{t\in \mathbb N\mid t\le3\})\vee (y>x)$
I've placed a question mark because it doesn't make sense, now $y$ is undefined for $(y>x)$. Not negating the $y\in$ leaves me with the same problem: $(y\in \{t\in \mathbb N\mid t\le3\})\vee (y>x)$ and it's probably wrong.
Maybe De Morgan could help: $(y\in \{t\in \mathbb N\mid t>3\})\wedge (y\le x)$ now I'm allowed to distribute $\forall$ so we get (not sure if it helps though):
$$\{x\in \mathbb R\mid\forall y(y\in \{t\in \mathbb N\mid t>3\})\wedge \forall y(y\le x) \}$$
I don't know how to continue from here, could be $\{x\in \mathbb R|x\ge 4\}$?
Any feedback would be appreciated.
EDIT: I think I made a mistake with De Morgan, it should be:
$$\{x\in \mathbb R\mid\exists y\left [(y\in \{t\in \mathbb N\mid t>3\})\wedge (y\le x)\right ] \} \overset {take: y=4}\equiv \{x\in \mathbb R|x\ge 4\}$$
You can write the main set as $$\{x \in \mathbb{R} : \forall y \in \{4,5,\ldots\} \quad y>x \}.$$ Then, it is clear that the main set is equal to $$\{x \in \mathbb{R} : x < 4\}.$$