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In the following let $\mathbf{V} = \bigcup_{\alpha \in \mathbf{ON}} V_\alpha$ denote the cumulative hierarchy. Let $\{\varphi_0, \dots, \varphi_n, \dots \}$ denote a list of all $ZF$ axioms. I am reading the following sentence:

"Given $n \in \omega$, the symbols "$\mathbf{V} \models \{\varphi_0, \dots, \varphi_n\}$" and "$\varphi_0 \land \dots \land \varphi_n$" stand for exactly the same thing." (Just/Weese, p 192)

I don't understand how this is true. The first expression says that $\varphi_i$ are all true in $\mathbf{V}$ given any valuation. The second expression seems to be a formula that may or may not be true but there is nothing saying that it is true in $\mathbf{V}$. Thank you for shedding light into my confusion.

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up vote 1 down vote accepted

In the presence of the axiom of Foundation/Regularity, $\mathbf V$ is exactly the class of all sets. Thus being "true in $\mathbf V$" is just the same as being true, period.

Technically, relativizing to $\mathbf V$ (which is what I think "$\mathbf V \vDash$" must mean here) is a no-op.

.. or put differently: The symbols "$\mathbf V\vDash\{\varphi_0,\ldots,\varphi_n\}$" is an instruction at the metalevel for you do do various manipulation that results in a wff in the language of set theory. The resulting formula contains no $\mathbf V$, no $\vDash$ and so forth, because these symbols are not in the language of set theory.

Similarly the symbols "$\phi_0\land\cdots\land\phi_n$" are an instruction for you to construct a certain wff in the language of set theory. The manipulations you do here are simpler, but at the least you need to do something about the "$\cdots$" which is not in the language, and unfold the "$\phi$"s to the formulas they denote.

The point is that these two sets of instructions result in wffs that are logically equivalent if only the formula (that results from unfolding) $\forall x.x\in\mathbf V$ is assumed to be true.

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But "$\varphi_0 \land \dots \land \varphi_n$ is true" is the same as saying "$\varphi_0 \land \dots \land \varphi_n$ is true in every model of $ZF$". On the other hand, "$\mathbf{V} \models \{\varphi_0, \dots, \varphi_n\}$" seems to be the same as saying "$\varphi_0 \land \dots \land \varphi_n$ is true in this model $\mathbf{V}$ of $ZF$". – Rudy the Reindeer Dec 18 '12 at 13:32
Also, does "$\varphi$" mean the same as "$\varphi$ is true"? – Rudy the Reindeer Dec 18 '12 at 13:33
@Matt: Why would "$\varphi$ is true" mean the same as "$\varphi$ is true in every model of ZF"? If it did, we couldn't speak of things being true in some models but not others. (Some of your other points may be addressed in my recent edit). – Henning Makholm Dec 18 '12 at 13:40
When we're speaking at the metalevel, saying "$\varphi$" itself just mentions a formula. Saying "$\varphi$ is true" makes a claim about some model, leaving it implicit which. It is fairly common to just say "$\phi$" when you mean "$\phi$ is true" when the context makes clear that a claim that can be right or wrong is expected. But notice that this still leaves it implicit which model/interpretation we're speaking of. – Henning Makholm Dec 18 '12 at 13:44
Then in the sentence I quote, they use "$\varphi_0 \land \dots \land \varphi_n$" to mean "$\varphi_0 \land \dots \land \varphi_n$ is true". Right? – Rudy the Reindeer Dec 18 '12 at 14:12

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