Predicate Logic - English Translation

Question

Can anyone please translate this Predicate Logic statement into English? The question is this:

Let $\mathbb{N}$ be the set on Natural numbers. Let $S=\{1,8,4\}$ and $T=\{6,8,11\}$.

Which of the following statements are correct?

$∃ x. ∀ y. (x ∈ T \land y ∈ T) \land (x ≤ y ∧ x ∈ S)$

I read it as:

There exists an $x$ for all $y, x$ is in the set $T$ and $y$ is in the set $T$ and there exists an $x$ that is less than or equal to $y$ and $x$ is in the set $S$.

The correct answer was False.

I interpreted this as meaning for every single value of $y$ there is a value of $x$ that is less than or equal to $y$ and every value of $x$ is also in the set $S$.

Is this correct?

Answer 1

Yes, it is false, indeed.

Witness that $8$ is the only element in the intersection of $S,T$, and that this is greater than $6$, an element in $T$.   Thus disproving by counterexample.

More so, witness that $1964$ is not in $T$; so not every $y$ is in $T$ even if there exists an $x$ in $S\cap T$.

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$∃x.∀y.(x∈T∧y∈T)∧(x≤y∧x∈S)$

"There is some $x$ such that $x$ is in $T$ and $S$ and every $y$ is in $S$ and at least as great as the $x$."

Answer 2

Don't re-use a quantifier - the $\exists x$ applies to the whole sentence. So the sentence says "there is an $x$ so that for every $y$, $x$ is in $T$ and $y$ is also in $T$ and $x$ is less than or equal to $y$ and $x$ is in $S$." So that means that $x$ has to be in both $S$ and $T$ and must be less than or equal to every $y$, and every $y$ is in $T$. But that clearly can't happen - not every number is in $T$.