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I've seen different definitions and am wondering what is the modern convention.

'$\mathcal{M}\vDash\phi(x)$'

Def. 1: $\mathcal{M}$ thinks $\phi(x)$ is true on some interpretation.

Def. 2: $\mathcal{M}$ thinks $\phi(x)$ is true on all interpretations.

'$\phi(x)\vDash\psi(x)$'

Def. 1: For every structure $\mathcal{M}$ and every interpretation $s$, if $\mathcal{M}\vDash\phi(x)[s]$, then $\mathcal{M}\vDash\psi(x)[s]$.

Def. 2: The same as $\forall x\phi(x)\vDash\forall x\psi(x)$ i.e. the formulas are understood as their universal closures.

IllogicalUser
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  • In my experience the answer in each case is "Definition 1" - usually. But note that this leads to a break with an earlier equivalence: we no longer have $$\varphi(x)\models\psi(x)\quad\iff\quad [\forall\mathcal{M}(\mathcal{M}\models\varphi(x)\iff\mathcal{M}\models\psi(x))].$$ For this reason I personally try to avoid both notations. – Noah Schweber Jun 01 '22 at 17:28
  • @NoahSchweber I am not sure why someone would want that equivalence. It doesn't make intuitive sense since the left hand side is a conditional, but the right hand side is a biconditional. – IllogicalUser Jun 01 '22 at 18:32
  • Sorry, that was a typo: the right-hand-side $\iff$ should have been an $\implies$. – Noah Schweber Jun 01 '22 at 18:50
  • For Question 2, see my answer here. For Question 1, I agree with everyone else here that the notation $M\models \varphi(x)$, when $\varphi(x)$ has free variables, should be avoided at all costs. – Alex Kruckman Jun 01 '22 at 19:04

1 Answers1

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'$\mathcal{M}\vDash\phi(x)$'

This is rarely used and myself I'd recommend not to use it. I would interpret it as $\mathcal{M}⊧ ∀x \phi(x)$.

'$\phi(x)\vDash\psi(x)$'

This is used more frequently (also as $\phi(x)\vdash\psi(x)$). It means that $\mathcal{M}\vDash∀x[\phi(x)\rightarrow\psi(x)$ for every $\mathcal{M}$.

Primo Petri
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  • Isn't it used all the time in model theory for talking about types? A set of formulas $p(x)$ is a partial type if there is a model $\mathcal{M}\vDash T\cup p(x)$. – IllogicalUser Jun 01 '22 at 18:23
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    I agree. The second one is very common, although mostly with $\vdash$ rather than $\models$ (also with types, not just formulas), while the first one, while technically correct, is awkward, not that useful, and (in conjuction with the other one), simply confusing. – tomasz Jun 01 '22 at 18:27
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    @T-Biconditional: That is incorrect. $p(x)$ is a partial type if there is an $M\models T\cup \exists x p(x)$. – tomasz Jun 01 '22 at 18:28