Does the empty model satisfy a contradiction?

Question

Suppose we are working in a formulation of first-order logic that allows empty models. Then, does the empty model satisfy the formula $(x=x \land \neg x=x)$? And if so, why doesn't the empty model satisfy all formulas, since it satisfies a contradiction? Note, the formula I am referring to is an open formula, it doesn't have a quantifier. I would like some clarification of this point.

Answer 1

The standard way to make sense of $\mathfrak{M}\models\varphi$ when $\varphi$ is a formula with free variables is to demand that $\mathfrak{M}$ satisfy every instantiation of $\varphi$; this is the same as $\mathfrak{M}$ satisfying the universal closure of $\varphi$. Under this definition, if we admit the empty structure we do indeed have $\emptyset\models(x=x)\wedge \neg(x=x)$.

However, it is no longer the case that a contradiction (in the sense of a formula gotten by substituting first-order formulas for propositional variables in an unsatisfiable propositional sentence) is explosive, as this example demonstrates. Variable-free contradictions are still explosive, of course. Basically, since we've changed our semantics we have to re-evaluate how our various notions relate to each other. And if we interpret "contradiction" as "unsatisfiable formula," then $(x=x)\wedge \neg(x=x)$ is no longer a contradiction ... as, again, this example demonstrates.

So there's no tension here at all, just a disconnect between two previously-equivalent properties arising because of a slightly broader semantics.

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And if that's not how you're defining "$\mathfrak{M}\models\varphi$" when $\varphi$ has free variables, how are you defining it?