need help with expressing uniqueness of elements that have the same property

Question

In class we were discussing how to translate the sentence there's a sucker born every minute .

The teacher showed us two examples

1: which is the wrong way

$\exists x [ sucker(x) \land \forall y(mintue(y) \rightarrow bornAt(x,y))]$

2: the correct way

$\forall y[minute(y) \rightarrow \exists x(sucker(x) \land bornAt(x,y))]$

I wanted to take my understanding further and make a wff that said there's a sucker that's born every minute and each sucker is also different from each other without paraphrasing the original sentence.

are any of these the correct way of translating it?

$\forall y[minute(y) \rightarrow \exists x(sucker(x) \land bornAt(x,y) \land \forall z (sucker(z) \rightarrow z \neq x))]$

$\forall y[minute(y) \rightarrow \exists x(sucker(x) \land bornAt(x,y) \land \lnot\exists z (sucker(z) \land z=x))]$

Answer 1

Neither of those express what we want them to.

$\forall z \mathop. (S(z) \to z \neq x)$ is equivalent to $\lnot \exists z \mathop. (S(z) \land z = x)$. Both mean all suckers $z$ are distinct from $x$ (where $x$ is a free variable denoting a sucker). Those statements are false when $z$ and $x$ refer to the same entity.

There are a few properties of suckers and when they're born that you might be talking about by saying that each sucker is different from the others. That could be taken to imply that there's a 1:1 mapping between suckers and minutes.

I'm interpreting the sentence as, instead, a consequence of observation that suckers are born at most once and therefore any sucker we find representing some minute can't be a representative of any other minute.

What we want to do is express two different facts about suckers:

1. $\forall y \in M \mathop. \exists x \in S \mathop. B(x, y)$ -- for every minute $y$, there's a sucker born at $y$.
2. $\forall y \in S \mathop. \exists_{\le 1} x \mathop. B(x, y)$ -- every sucker is born at most once.

This is equivalent to:

1. $\forall y \mathop. ( M(y) \to \exists x \mathop. S(x) \land B(x, y))$
2. $\forall y \mathop. (S(y) \to \forall a \forall b \mathop. (B(y, a) \land B(y, b) \to a = b)) $