Formula in upper Brackets Notation $\ulcorner \varphi \urcorner$: Meaning and Usage

Question

I have some understanding problems with 2nd paragraph in following answer by spaceisdarkgreen:

[...] Note that from this perspective, when we write, e.g. $M\models \exists ! x\varphi(x)$ or $M\models \sf ZFC$ in explaining what $\omega_M$ is, this is an abuse of notation. Here $\varphi$ is not a formula, but rather an analogue of a formula formalized in our background theory (and similarly the reference to "$\sf ZFC$" is to a formalized version of this set of formulas). It just so happens that our background theory is $\sf ZFC$ (or something close enough to it) so the formula $\varphi$ in our background language has a corresponding formal object $\ulcorner \varphi\urcorner$ that our background theory can talk about $M$ satisfying.

Fistly, The setting is that we have a well-formed formula $\varphi$ in language of a theory $T$ - here $T=ZFC$; we assume that the proposition $ \exists ! x\varphi(x)$ is provable in $T$ and $M$ is some picked model of $T$, so we neccessarily have $M\models \exists ! x\varphi(x)$.

Now spaceisdarkgreen remarked that writing $M\models \exists ! x\varphi(x)$ is strictly speaking ill parsed in the sense of it admits abusion of notation.

Question: Could somebody elaborate in which sense this term $M\models \exists ! x\varphi(x)$ actually admits abusion of notation and what is actually meant by $\ulcorner \varphi\urcorner$ and how is it precisely related to original $\varphi$?
Would phrasing it as $M\models \exists ! x \ulcorner \varphi(x) \urcorner$ instead eliminate this adressed issue with ill notation/ notation abusion?

A guess/ my considerations: So far I understand it the issue with abusion of notation is that $M\models \exists ! x\varphi(x)$ is intrinsically a metastatement about theory $T$ and its model $M$, and in metatheory of $T$ (...spaceisdarkgreen called it "background theory, but guess that that's the same as metatheory, right?) the language is a priori different from the underlying language of $T$, so the metatheory would simply a priori not be able to recognize $\varphi$ as formula.
Is this precisely the main reason why $M\models \exists ! x\varphi(x)$ is regarded to abuse notation? Or is the reason a different one?

Now in specific situation above spaceisdarkgreen noticed that we are lucky because in this specific situation the background theory is also $ZFC$, so have same alphabets and so $\varphi$ would be well-formulatable also in background theory language.

Now my guess is that the notion $\ulcorner \varphi\urcorner$ serves only to distinguish/explicitly indicate when we regard the formula in $T$ or in background/ metalanguage, and that's it?
Or do I misunderstand what is actually the honest reason to consider the notation $\ulcorner \varphi\urcorner$ distinguishing it from just $\varphi$ without brackets?

Answer 1

See Quasi-quotation.

In a nutshell, $\varphi$ is used math logic as a schematic variable: we use it to refer to a generic formula. But when we write "$\varphi$", what we are referring to, in the current usage of quotation symbols, is not the name of a generic formula but the name of a Greek letter.

The issue is that substitution inside usual quotation symbols is not allowed; see Willard Van Orman Quine, Mathematical logic (Harvard UP, rev.ed 1951), page 33.

In the above paragraph the issue seems to me a little bit convoluted: in the formal language of set theory a formula is an expression like $(x ∈ ∅)$, while $\vDash$ is not a symbol of the language.

The expression $M \vDash ¬(x ∈ ∅)$ is an expression of the meta-language and says that... Also $\varphi$ is not part of the language but it is a schematic symbol used in the meta-language (the part of mathematics where we speak of axioms and theorems) to refer to a "generic" formula of the formal language.

Thus, to make a silly example, the meta-language expression $\mathsf {ZFC} \vdash \varphi$ is false, because $\varphi$ stands for a formula whatever and, if set theory is consistent it does not prove every formula.

In conclusion, see the discussion in Quine: to be formally accurate, an expression like $φ \lor ψ$ is formally wrong, because it is a mix of language and metalanguage.