# Precise definition of valid argument using model theory

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?

• 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. 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. 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