# First order logic model: why not an ordered triplet?

I've started studying first order logics semantics. By and large, a FOL model is definied by an ordered couple, $$\langle D, \mathcal{I}\rangle$$, where $$D$$ is the domain and $$\mathcal{I}$$ is an interpretation function mapping individual constants to elements of $$D$$, and predicate constants to $$D$$'s subsets.

But when it comes to evaluate open formulae we also need a variable assignment function, $$g$$, that maps every individual variable to an element of $$D$$.

This is why I wonder: why is a FOL-model defined by an ordered pair, $$\langle D, \mathcal{I}\rangle$$, and not – as I would expect – by an ordered triplet $$\langle D, \mathcal{I}, g\rangle$$?

Maybe the question may sound silly to those who are familiar with the subject, but since I am just getting into this kind of study I need a conceptual clarification.

• Variables wouldn’t be very variable if you fixed their values. Commented Aug 18, 2023 at 18:38

The ordered pair $$$$ is a structure, a model of a sentence P is a structure that satisfies P. And indeed the typical notion of satisfaction takes into account variable assignments.

As to why models do not really need a variable assignment: recall that a model satisfies sentences - ie, closed formulas. Hence, variables appearing in the sentences are bound, and their meaning is in a sense determined purely via the quantifiers. in other words, we only need to know that something holds true for all assignments or for one assignment (forall, exists resp).

Example: in any model of a group, say the integers, associativity holds. That is, for any assignment of x, y, z we have x(yz) = (xy)z.

That's actually a great question! I believe the ordered triplet definition would work perfectly fine. That is, you could develop FOL by considering models that include the coordinate $$g$$ assigning meaning to all free variables. In fact, there would be no real difference between constants and variables in such a system. See, for example, my related question:

First-order logic where constants play the role of variables

So why don't we usually include the variable assignment function $$g$$ in a model? The general reason is that in first-order logic, we are ultimately interested in the meaning of sentences, which are formulas where there are no free variables. For example, we ultimately want to consider sentences like $$\forall x. \exists y. x > y$$ and whether they are true in the model -- we don't want to consider unquantified formulas like $$\forall x. x > y$$ to be meaningful -- or at least, they only have meaning when bound by a particular variable assignment to $$y$$. The fact that models do not include a variable assignment function $$g$$ is an artifact of this philosophy -- this ensures that we don't make a mistake of conflating constants in the model (such as the constant $$0$$ or $$1$$) with variables that are assigned temporarily or locally in the context of a bounded quantification (like $$x$$ and $$y$$ above).

Let me know if that helps!

• Let's look at a specific model in the theory of groups, say $(\Bbb{Z}, \{0 \mapsto 0, {+}\mapsto (i, j) \mapsto i + j\})$. What have variable assignments got to do with the model? Commented Aug 18, 2023 at 22:24