Predicate Logic - Why is this not sufficient?

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Let N be the set on Natural numbers. Let S={2,6,14} and T={6,8,11}.

Question: The intersection of S and T is nonempty

The correct answer is this:

$∃ x : x ∈ S. ∃ y : y ∈ T. x=y$

My answer is this:

$∃ x : x ∈ S ∧ x ∈ T$

Is my answer a sufficient way of saying the value x is in the intersection of S and T?

If not, why does the correct answer refer to two variables, x and y? Can x only represent the values in one set, even when they intersect?

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Your answer is correct ... if you add some parentheses:

$\exists x (x \in S \land x \in T)$

This is logically equivalent to the provided answer, and so no, you don't need a different variable just because you have different sets; it is the same object that is in the intersection, so you can indeed do exactly what you do.

Finally, your statement is not quite saying 'x is the intersection of S and T' (because for different S and T they may have more than 1 element in common), but it is saying that 'x is in the intersection of S and T', and hence that the intersection is not empty.