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So, if $T$ is a theory in a first order language $\mathcal L$, I thought a model for $T$ is

a set $M$ with interpretations for all the constant, function and relation symbols of $\mathcal L$, in which all statements in $T$ are true.

But recently, I had someone calling a model

a function $\beta$ that

  • assigns a truth value to each $\mathcal L$-statement,
  • assigns "true" to all statements in $T$, and
  • is compatible w.r.t. the usual inference rules, e.g. $\beta(\phi\land\psi)$ gives "true" if and only if $\beta(\phi)$ and $\beta(\psi)$ both give "true".

Question:

  • Is the latter also called a model of $T$?
  • Are these definitions equivalent, and if so, are they equivalent in an obvious way?

Some thoughts

So clearly, each "set-model" $M$ given such a "function-model" $\beta$ by defining

$$(*)\quad \beta(\phi)=\text{"true"}\;\Longleftrightarrow \; M\Vdash \phi$$

Also, if $T$ is inconsistent by proving $\phi\land \neg\phi$, then we cannot choose a value for $\beta(\phi)$, and so no such "function-model" exists.

The other direction seems non-obvious. I can imagine that the equivalence follows from Gödel's completeness theorem, but I am uncertain. Is it true that for every choice of such a "function-model" $\beta$ there is a "set-model" $M$ so that $(*)$ holds?

M. Winter
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1 Answers1

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No, that second definition is totally unsatisfactory (for first-order logic anyways - it's the correct definition for propositional logic).

First, let me rephrase it slightly. By thinking about $\beta^{-1}(\{\top\})$ instead of $\beta$ itself, we see that it's just a definition of a maximal consistent theory containing $T$. This is a bit easier to think about in my opinion, so I'll do so. Maximal consistent sets (even with additional bells and whistles - see below) are definitely not the same things as a models, for two big reasons:

  • Until we prove the completeness theorem, we don't know that a maximal consistent $S$ has a model.

  • More fundamentally, maximal consistent theories do not have unique models up to isomorphism (unless they're the theories of finite structures), per the compactness theorem. Put another way, isomorphism is finer than elementary equivalence (and indeed inasmuch as model theory studies complete theories - which is most of the time, really - it's leaning on the richness which can occur within a single elementary equivalence class).

Now to be fair, if you really really want to avoid talking about sets you could look at expansions of the language instead and get something kinda-reasonable. Specifically, a "syntax-only model" of a theory $T$ in a language $\mathcal{L}$ could be defined as a pair $(S, \hat{\mathcal{L}})$ where:

  • $\hat{\mathcal{L}}\supseteq \mathcal{L}$,

  • $S$ is a maximal consistent $\hat{\mathcal{L}}$-theory containing $T$, and

  • $S$ has the witness property: whenever $\exists x_1,...,x_n\varphi(x_1,...,x_n)$ is a sentence in $S$, there are closed $\hat{\mathcal{L}}$-terms $t_1,...,t_n$ such that $S\vdash\varphi(t_1,...,t_n)$. (I've said a bit about the importance of the witness property here which may be relevant.)

Such a syntax-only model has a naturally associated structure in the usual sense which (has a reduct which) is indeed a model of $T$. Moreover, every consistent theory has a syntax-only model; this is really what Henkin's argument shows. So we can get something via a syntax-only approach (but we have to add the witness property, which isn't stated in your second definition - or at best is unclearly implicit in its final buletpoint).

However, not all models of $T$ will occur this way. In particular, any model of $T$ with cardinality $>\vert T\vert+\aleph_0$ will not be so represented. So again, lots of model-theoretic topics vanish if we make this shift (or at least require tedious circumlocutions).

Noah Schweber
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