$M \not \models \phi$ vs $M \models \neg \phi$

Question

Let $\phi$ be a First Order formula. If $M \not\models \phi$ does it mean that $M \models \neg \phi$? Why?

Answer 1

Yes. If $M$ is a structure, then "$M\models\neg\varphi$" means by definition "$M\not\models\varphi$."

However, note that things get more complicated at the level of theories. A theory $\Gamma$ - that is, a set of sentences - is said to satisfy a sentence $\varphi$ if $\varphi$ is true whenever $\Gamma$ is true; that is, $$\mbox{for all structures $M$, if $M\models\Gamma$ then $M\models\varphi$.}$$ (Here "$M\models\Gamma$" is shorthand for "$M\models\psi$ for every $\psi\in\Gamma$.") If $\Gamma$ satisfies $\varphi$, we write (perhaps confusingly) "$\Gamma\models\varphi$."

Now it is no longer true that "$\models\neg$" is the same as "$\not\models$"! This is because theories don't have to decide everything, whereas models do. For example, if we take $\Gamma$ to be the set of group axioms, and let $\varphi$ be the sentence $\forall x\forall y(x*y=y*x)$, then

• $\Gamma\not\models\varphi$, since there are nonabelian groups; but also

• $\Gamma\not\models\neg\varphi$, since there are abelian groups.

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Note: there is nothing special here about first-order logic! Second-order logic, infinitary logic, logic with a cardinality quantifier, etc. also behave the same way. Basically, any logic founded on classical propositional logic (two truth values, excluded middle, etc.) will work this way, since the truth definition for negation will be "$M\models\neg\varphi\iff M\not\models\varphi$". Note that things will be different if we look at logics founded on intuitionistic propositional logic, but that's a whole other can of worms.