In philosophy 101, I learned that a valid argument is any argument that satisfies this property: if all of its premises are true, then its conclusion must hold true.

Now, I am taking a class on metalogic where I learn about first-order model theory. I wonder how the definition of valid argument can be rewritten in a more precise manner in terms of model theory.

One possible way I can rewrite the definition is to first define an argument as a pair: Γ and A, where Γ is a set of sentences (premises) and A is a sentence (conclusion). Then I define an argument (Γ, A) to be valid if and only if for any structure M with M⊨Γ, M⊨A.

I wonder how this precise definition can be put into practice. Consider this simple example of an invalid argument: (Γ, A)=({p→q, q}, p). How can I use the definition to show it is invalid?

I am not sure what is the appropriate first-order language for this context. For now, let's assume that our language L consists two constant symbols: p and q.

In order to show that (Γ, A) is invalid, I can demonstrate with a L-structure M, which has domain {⊤, ⊥} and interpretation: p^M=⊥, q^M=⊤ (I am not sure if symbols ⊤, ⊥ are allowed to be in the domain). With this structure, we have M⊨Γ since M⊨p→q and M⊨q, but M⊨A is false, so (Γ, A) is invalid. Is that correct?

  • 2
    Yes, you have it right. In model-theoretic terms, an argument is valid if every model of the premises is a model of the conclusion. In your example, you only need the propositional calculus and you are correct that there is a countermodel when q is true and p is false.
    – Bumble
    Feb 3 at 3:18
  • See IEP, Model-Theoretic Conceptions of Logical Consequence. However, you do not really need domains if you only consider propositional formulas. Their models just collapse to truth tables, each line of the truth table that assigns ⊤or ⊥ to each propositional letter is a "model". Domains are not for propositional letters but for variables in predicates. Variables run over elements in the domain when the truth of quantified statements is evaluated.
    – Conifold
    Feb 3 at 4:30
  • You're absolutely correct. And yes, you don't need the domains, as Conifold said. And models without domains are called: possible worlds!
    – user71009
    Feb 3 at 6:27

1 Answer 1


@Conifold is correct to say that a model of only propositional formulas is essentially just a truth table. However, if you insist on taking it to the first order level, a standard treatment follows with a multi-sorted logic.

In particular, our vocabular now consists of a set of sorts, with relations of type (sort_1, ..., sort_n). Functions generalize in a similar manner. A structure U is then a family of sets, one for each sort, equipped with an interpretation for each relation symbol R that is interpreted as a subset of Prod_i (sort_i^U). If the arity of R is 0, then Prod_i (sort_i^U) is the empty product, ie a set (by typical set -theoretic considerations) with one element. Call this set {o}. Likewise with functions. In particular, if the relation is of arity zero, it is a subset of {o}, ie either the empty set which we interpret as false or {o} which we interpret as true. In this manner, we can add propositional formulas to our language and give them a semantics. In other words, almost exactly as you have done, but with extra sorts.

see https://plato.stanford.edu/entries/logic-many-sorted/ for more.

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