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.

Thak you in advance

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

2 Answers 2


The ordered pair $<D, I>$ 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!

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    $\begingroup$ 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? $\endgroup$
    – Rob Arthan
    Commented Aug 18, 2023 at 22:24

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