Is there a name for models whose every element is named by (one or more) variable-free terms?

Question

Let $T$ denote a first-order theory. Is there a name for those models $M$ of $T$ such that for all $x \in M$, there is a variable-free term in the language of $T$ whose interpretation under $M$ is $x$?

For example, let $T$ denote the first order theory in the language of $\{S,0\}$ generated by the axioms:

1. $Sx \neq 0$
2. $x \neq y \rightarrow Sx \neq Sy$

Then I'm pretty sure that any model $M$ of $T$ having the aforementioned property will be isomorphic to $\mathbb{N}.$

Answer 1

These models are called term models in Wolfgang Rautenberg's A Concise Introduction to Mathematical Logic. In Section 4.1 of this book, you find some sufficient conditions for a theory to have a term model. I am not sure how widely used this terminology is, but I like this name.

Answer 2

I have never heard of a terminology for this kind of models, but there is a well known terminology for theories that admit a model like that.

First, you are absolutely right about your last comment, and it easily generalizes: if $T$ is a complete theory and $\mathcal{M}$, $\mathcal{N}$ is a pair of models of $T$ with the aforementioned property than $\mathcal{M}$ and $\mathcal{N}$ are isomorphic: just take $t^\mathcal{M}$ to $t^\mathcal{N}$, for every constant term $t$. This map is always well defined and partial elementary. The fact that it is surjective and defined everywhere follows from the hypothesis on our two models.

Edit: the theory in question is not complete. You can't tell if there is an element other than $0$ with no predecessor. It is still true that every pair of models where every element is a constant are isomorphic. In any model the elements $0, S0, S00, \ldots$ are all distinct because $S$ is injective. These are the only constant terms, so the obvious map between two models where every element is of this form is an isomorphism.

My answer applies only in the case of a complete theory.

Let $T$ be a complete theory. If $T$ has the property that for every formula $\phi(x)$ there is a consant term $c$ in the language such that $T \models \exists x \phi(x) \to \phi(c)$, then $T$ is said to have the Henkin witness property. In this case, $T$ has a canonical model with the property you mention. This appears in the standard proof that every consistent set of formulas has a model, for example in Marker's book. The canonical model will have elements of the form $t$, for $t$ a constant term in the language, quotiented out by the relation $t \equiv s$ if $T \models t = s$.

Conversely, assume that $T$ is a complete theory and $\mathcal{M}$ is a model with the property you mention. Then I claim that $T$ has the Henkin witness property. To see this, let $\phi(x)$ be a formula and suppose that $T \models \exists x \phi(x)$. Then there s some $m \in \mathcal{M}$ such that $\mathcal{M} \models \phi (m)$. But $m$ has form $c^\mathcal{M}$ for some constant term $c$ in the language; so $T \models \exists x \phi(x) \to \phi(c)$.

So I would say that a model with this propety is the canonical model of a theory with the Henkin property (the theory is $\text{Th}(\mathcal{M})$)

By the way, note that it is in general completely hopeless to ask that all the models of a theory be like this: if there are infinite models then there are models of arbitrary large cardinality, and it is impossible that every element in a big model is the interpretation of a constant term because there are simply not enough terms.