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Just as an example, in some places the Axiom Schema of Comprehension is formulated with a free variable for example Kunen:

Axiom 3. Comprehension Scheme. For each formula, φ, without y free, $\exists y \forall x (x \in y \longleftrightarrow x \in z \land \phi(x))$

In other places such as Wikipedia and Hrbacek, Jech formulates it without free variables:

The Axiom Schema of Comprehension: Let P(x) be a property of x. For any set A, there is a set B such that x ∈ B if and only if x ∈ A and P(x).

There is a difference in what can be deduced inside ZFC (I mean without using model theory or semantic resources) between

$\exists y \forall x (x \in y \longleftrightarrow x \in z \land \phi(x))$

and

$\forall z \exists y \forall x (x \in y \longleftrightarrow x \in z \land \phi(x))$ ?

In general, What is lost if we completely exclude free variables from ZFC?

I know that "There is nothing specific in 𝖹𝖥𝖢 regarding free variables" but we can build a ZFC theory where all formulas are closed. What limitation may have that theory?

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    $\begingroup$ No difference; in mathematics, when we asserts a formula, like e.g $x+y=y+x$, we assume that the free vars are implicitly universally quantified. Thus, Axiom of Separation $∃y∀x(x∈y ⟷ (x∈z \land \varphi))$ must be read as $∀z∃y∀x(x∈y⟷(x∈z \land \varphi))$. $\endgroup$ Oct 4 '18 at 10:51
  • $\begingroup$ @bof It is not an axiom it is an axiom scheme. That is an axiom for each possible formula φ. $\endgroup$ Oct 4 '18 at 10:54
  • $\begingroup$ @bof Right. I will edit it. $\endgroup$ Oct 4 '18 at 11:05
  • $\begingroup$ I hope you do appreciate that you need to use formulas with free variables to carry out proofs in the standard presentations of first-order logic. However the inputs (axioms) and outputs (theorems) of proofs can always have their free variables quantified without making any significant difference. $\endgroup$
    – Rob Arthan
    Oct 4 '18 at 19:43
  • $\begingroup$ @RobArthan No, I don't. Can you explain it? Can you give an example? $\endgroup$ Oct 5 '18 at 3:48
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In fact Kunen's book has all the axioms as closed formulas: in both Axiom 3 (Comprehension) and Axiom 6 (Replacement) it says "the universal closure of the following is an axiom"; the other axioms are all single closed formulas.

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    $\begingroup$ So there is not any formal lost getting rid free variable from ZFC, it is just a matter of presentation. Right? $\endgroup$ Oct 4 '18 at 11:02
  • $\begingroup$ Indeed, in Mendelson's Introduction to Mathematical Logic both forms would be formally deducible from each other. $\endgroup$
    – hartkp
    Oct 4 '18 at 11:07

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