Is the following second-order statement true or false? (Assume $P$ is a binary predicate statement which is always defined for two members of $S\neq\emptyset$.)
$$\forall P\left( x,y \right)\forall S \neq \emptyset \Bigl[ \left[ \exists \left( a,b \right)\! \in\! S^2\, P \left( a,b \right) \middle] \equiv \middle[ \exists \left\{ a,b \right\}\! \subseteq\! S\, P \left( a,b \right) \right]\Bigr]$$
If the statement is true, does it still hold if the two $\exists$ quantifiers were replaced with $\exists!$ or $\forall$ ? Otherwise, if the statement is false, please provide a counterexample involving a specific $S$ and $P$ and explain why the distinction between $\exists\left(a,b\right)\in P^2$ and $\exists\left\{a,b\right\}\subseteq P$ matters.
For some context, in my (admittedly quite limited - I'm an undergrad) experience with formal logic I've always seen one of two general forms for a quantified predicate statement, where the predicate $P$ takes two arguments $a$ and $b$, both belonging to the same set $S$... $$\exists a\! \in\! S \exists b\! \in\! S\; P \left(a,b\right) \;\;\; \mathrm{or}\;\;\;\; \exists \left( a,b \right)\! \in\! S^2\; P \left( a,b \right)$$ ... with the former generally being avoided when the domain of $a$ and $b$ needs to be explicitly stated as I have done here.
That said, I don't think I've ever seen a statement written similarly to... $$\exists \left\{ a,b \right\} \subseteq S \; P \left( a,b \right)$$ ... unless the predicate $P \left( a,b \right)$ only deals with the set $\left\{ a,b \right\}$ directly, in which case writing it as a unary predicate $P \left(\left\{ a,b \right\}\right)$ might be more appropriate to the context.
Wondering why $\exists \left( a,b \right)\! \in\! S^2\; P \left( a,b \right)$ seems to be preferred over $\exists \left\{ a,b \right\} \subseteq S \; P \left( a,b \right)$ has led me down this rabbit hole. I can't think of any example where replacing the former with the latter changes the truth value of the predicate statement, only examples where using the latter feels "out of place" in context. For example, in a situation where $P \left( a,b \right) \Longrightarrow a=b$ is known to be true, any set $\left\{ a,b \right\}$ witnessing $\exists \left\{ a,b \right\} \subseteq S \; P \left( a,b \right)$ must have only one distinct member, but this does not mean that a witness set cannot exist.
I've concluded that the most likely explanation to my never seeing $\exists \left\{ a,b \right\} \subseteq S \; P \left( a,b \right)$ in the wild is that a number of "soft" factors related to my relative inexperience with formal logic in broader contexts are to blame, not that the two examples I've given are not logically equivalent. For instance, finding the truth value of $f\left(a\right)=b$ is an easily accessible example for an intro to discrete mathematics course, and using an ordered pair $\left( a,b \right)$ makes inherent sense in that context. However, using a set $\left\{ a,b \right\}$ could allow $\left|\left\{ a,b \right\}\right|=1$ as mentioned above. Obviously this is a speculative guess so it can't be empirically confirmed, but if there is some case where my two example statements are non-equivalent (i.e. the second-order statement I am asking about overall is false), I would have an objective explanation to my pondering.