I (and possibly others) naively thought of $\forall$ as an infinite conjunction (only vaguely aware of others saying to avoid this interpretation of the $\forall$ sign). Years later I'm confronting this belief. I've read over Derek Elkins' blog post on the matter. I only partially probably have the logical prerequisites to approach the problem, but I'm first trying to state rigorously what I want to say. In what follows, infinitary logic is required as otherwise you can't talk about infinite conjunction.
Progress towards defining the problem so far: The first issue with comparing $\forall xP(x)$ and $\bigwedge_{i\in\mathcal{I}}\phi_i$ is that the $\phi_i$ have nothing to do with $P$. So I can narrow my question: I want $\phi_i$ to be related to $P$. But something like $\bigwedge_{i\in\mathcal{I}}\forall xP(x)$ would be silly. This made me realize: what I really need is a set $\mathcal{U}$ of constant symbols, and for any $c\in\mathcal{U}$ to let $P(c)$ be the resulting formula you get when free occurrences of $x$ in $P(x)$ are substituted for $c$. Now we're closer to stating a comparison that makes sense: what is the difference between $\forall xP(x)$ and $\bigwedge_{x\in\mathcal{U}}P(x)$? Moreover, since we are firstly intrested in the $P(x)$ from finitary logic, we'll say that $P(x)$ is definable in finitary logic (even though our background is infinitary logic). Furthermore, for simplicity, right now I'm only interested in $P(x)$ being a unary predicate - or you can think of it as being a well-formed formula of finitary logic with precisely $1$ free variable. The remaining choice is how we define $\mathcal{U}$.
Decision 1: Are constants in the set of constant symbols only given by extensions by definitions or can they also be created by extension by function names? See here and here for the respective differences. Essentially, extension by names doesn't require uniqueness whereas an extension by definitions does. If we allow a constant $c$ of $\mathcal{U}$ to be produced by extensions by names, then $\forall xP(x)\leftrightarrow\bigwedge_{x\in\mathcal{U}}P(x)$ holds because we can create a constant with only the defining axiom $c=c$, and since $c$ has essentially no properties particular to it, if we prove it for that constant $c$ then we prove the statement $\forall xP(x)$. So perhaps we want to define a constant symbol as degenerate if its defining axiom is satisfied by everything, i.e. if $Q(c)$ is the defining axiom of a constant $c$ but $\forall xQ(x)$ holds, then $c$ is a degenerate constant symbol. I am interested in what happens if we exclude degenerate constant symbols from $\mathcal{U}$, but it is not intuitive to me if it really makes sense to forbid all constant symbols that came from extension by names from being in $\mathcal{U}$.
Decision 2: If we only include $c$ in $\mathcal{U}$ where $c$ is borne of an extension by a definition, since we are in infinitary logic do we allow defining axioms to be infinitely long? If not, then the set of definable constants $\mathcal{U}$ is countable. Surely, if someone could affirm me, $\forall xP(x)\leftrightarrow\bigwedge_{x\in\mathcal{U}}P(x)$ does not hold in this case because for most semantics $\bigwedge_{x\in\mathcal{U}}P(x)$ ends up representing what's true only for definable elements of the model, which may be different from $\forall xP(x)$ which captures undefinable elements of the model.
Decision 3: If we only select constants borne of extension by definitions to belong to $\mathcal{U}$ and we allow infinitely long defining axioms for the elements $c$ of $\mathcal{U}$, is there a canonical way to do this?
Summary: In trying to make an infinite conjunction that is equivalent to $\forall xP(x)$, I don't know whether to include constant symbols in $\mathcal{U}$ that are only born of extension by names, and whether to allow infinitely long defining axioms for the constants in $\mathcal{U}$ and if so whether there is a canonical way to produce these infinitely long definitions. Does $\forall xP(x)\leftrightarrow\bigwedge_{x\in\mathcal{U}}P(x)$ hold for any of these variants?
Edits to explain how this is not a duplicate of: Are quantifiers only required because of infinite proposition chains?
The first answer provides a justification of why formulas need to be finite. I understand that, and that at some point in your metatheory using infinite conjunction doesn't make sense. But that's not what I'm asking. I'm asking if there is metatheoretical justification to treat the two as equivalent.
The first two answers suppose that an infinitary conjunction is restrictive of your domain. But in this question, I have proposed several possible definitions for infinite conjunction. The first answer stated that infinite conjunction restricts you to a definable domain. I understand how that is restrictive, but metamath is complicated and I'm like someone to confirm that in the case of my problem. But the real point is that in this question I'm defining multiple other possible, different definitions of infinite conjunction which are not clearly addressed in any of the answers (at least not clearly to me anyways). As for the second answer, I do not yet know what $\bigwedge_{x\in\text{dom}(\mathcal{M})}P(x)$ means. After all, don't we need to turn elements of $\text{dom}(\mathcal{M})$ into syntactic objects each with property that they have before we can work with them in a proof? How to use infinitary logic to give each element of your domain a defining axiom seems highly nontrivial and I can't begin to wrap my head around it.
The third answer mentions non-constructive proofs, but I've never seen a clear definition of what is or isn't constructive in math.
The nature of the proposed duplicate is different. It was asked by someone who did not even understand how to formulate their question properly so a simple answer sufficed. Half the point of my question is trying to figure out what the right formulation of the question is, so naturally my question requires a different answer.
