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.
Does the empty model satisfy a contradiction
first-order-logicmodel-theory
Related Solutions
We agree that an interpretation is a model of a formula $\varphi$ if the formula is TRUE in that interpretation (i.e. if the interpretation satisfies the formula).
The same for a set $\Gamma$ of formulas.
Consider now a theory $T$ with the collection $\Gamma_T$ of its axioms.
"A model of the theory is an interpretation that satisfies all the axioms of the theory."
So far, nothing new : the collection of axioms of the theory $T$ is a set of formulas. Thus, a model of the theory is an interpretation that satisfies all those formulas.
Consider for example first order arithmetic, i.e. the f-o version of Peano axioms.
We can call the collection of f-o Peano axioms with $\mathsf {PA}$.
We can prove from it the usual arithmetical laws (or theorems), like e.g. : $1+1=2$.
In symbols, we have : $\mathsf {PA} \vdash (1+1=2)$.
By soundness of the predicate calculus, we have : $\mathsf {PA} \vDash (1+1=2)$.
And again, this is consistent with the above definitions :
the arithemetical theorem $1+1=2$ is a logical consequence of the (first-order) arithmetical axioms, i.e. it is TRUE in every interpretation that satisfies the collection $\mathsf {PA}$ of arithmetical axioms.
In the answer to your previous post we have seen that an interpretation for propositional logic is :
an assignment $v : \text{At} \to \{ \text T, \text F \}$ such that, e.g. $v(p_0)= \text T, v(p_1)= \text F$, etc.
In the case of first-order language an interpretation needs a domain of "objects", like e.g. the set $\mathbb N$ of natural numbers.
Of course, different FOL theories need different domains, while in propositional logic we have only one domain: the boolean one $\{ \text T, \text F \}$.
And then we have to specify how to interpret the new elements of the language (in addition to the connectives) : quantifiers, individual constants, variables, predicate and function symbols.
Having done this, we have to complete our semantics for FOL with the definition of :
satisfaction relation, model, valid formula, logical consequence.
Think about how $\models$ is defined in the first place: by definition, $\mathcal{A}\models\neg\varphi$ iff $\mathcal{A}\not\models\varphi$. So excluded middle in the metatheory prevents the situation you're asking about: to get $\mathcal{A}\models\varphi$ and $\mathcal{A}\models\neg\varphi$ we'd need to have both $\mathcal{A}\models\varphi$ and $\neg(\mathcal{A}\models\varphi) $, which is impossible.
(And if that's not how you're defining $\models$, then how are you defining it?)
Best Answer
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.
And if that's not how you're defining "$\mathfrak{M}\models\varphi$" when $\varphi$ has free variables, how are you defining it?