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I'm reading a text on Set Theory which states that any formula, say $\phi$, is ultimately built up from atomic sentences of form $x \in y$ and $x = y$ via the logical connectives.

So then my question is as follows: suppose in reading this text I come upon a formula of form $\phi(x,y,z)$. Am I to interpret this formula as ultimately a claim about membership and equality? If $x,y,z$ are fixed, is $\phi(x,y,z)$ just really the claim $(x,y,z,1) \in \phi$ where $\phi$ is interpreted as a predicate function?

If this post isn't clear, here is really the question I'm getting at: how am I supposed to interpret statements like $\phi(x,y,z)$ when reading a book concerning axiomatic set theory written in the language of first-order logic?

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When you meet $\phi(x,y,z)$ then you know two things:

  1. $\phi$ is a formula, in this case written in the language of set theory which only includes $\in$ (and $=$ which is part of the logical symbols).
  2. $x,y,z$ denote the free variable which may (and usually do) occur in $\phi$.

This means that $\phi$ can be something like $x\in y\lor x=z$. But it also means that $\phi$ can be something much much much more complicated: $$(\exists w(x\in w\rightarrow y\notin z)\lor\forall w(y\in w\rightarrow z\in x))\rightarrow\exists w\forall u(u\in x\leftrightarrow x\in w\land y\in u)$$

These formulas come in two common flavors: as a general statement whose content doesn't really matter for us right now, we may be interested in its structure (in particular in the quantifiers it has inside), but not in its content; and sometimes it will be a statement similar to "Let $\phi(x,y,z)$ be a formula saying that $x$ is an ordinal, and $y$ is a bijection from $x$ onto $z$", in which case this is exactly how you should understand it.

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  • $\begingroup$ Your last paragraph starts a bit awkward (there is no counterpart to "either"). (On second reading that's not entirely true, but it's not really an archetype of perfect formulation.) $\endgroup$ – Lord_Farin Jun 6 '13 at 15:34
  • $\begingroup$ I'm just really tired. I'll re-read it, and correct what needs to be corrected in a few hours. $\endgroup$ – Asaf Karagila Jun 6 '13 at 15:38

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