Kees Doets's definitions of logical consequence

everyone, I'm reading Kees Doets's Basic Model Theory (which is freely and legally downloadable from https://web.stanford.edu/group/cslipublications/cslipublications/Online/doets-basic-model-theory.pdf). There are two different definitions of logical consequence in it. The official one ($$\vDash$$) is on p.6:

$$\Sigma$$ is a set of formulas, the notation $$\Sigma \vDash \phi$$ -- $$\phi$$ follows logically from $$\Sigma$$ -- is used in the case that $$\phi$$ is satisfied by an assignment in a model whenever all formulas of $$\Sigma$$ are.

The other, 'unofficial' one ($$\vDash ^*$$), is in Exercise 8, which is on p. 7:

Sometimes, logical consequence is defined by: $$\Sigma \vDash ^* \phi$$ iff $$\phi$$ is true in every model of $$\Sigma$$.

It bothers me how these two definitions are different. Indeed, this is exactly what Exercise 8 asks:

Show that if $$\Sigma \vDash \phi$$, then $$\Sigma \vDash ^* \phi$$, and give an example showing that the converse implication can fail. Show that if all elements of $$\Sigma$$ are sentences, then $$\Sigma \vDash \phi$$ iff $$\Sigma \vDash ^* \phi$$.

Let me say what I think about the first definition of logical consequence, i.e. $$\vDash$$. It seems to me that

$$\phi$$ is satisfied by an assignment in a model whenever all formulas of $$\Sigma$$ are.

just means

whenever all formulas of $$\Sigma$$ are satisfied by an assignment in a model, $$\phi$$ is satisfied by the same assignment in the same model.

which, appears to mean

for all models $$\mathcal{A}$$, for all assignments $$\alpha$$, for all $$\psi \in \Sigma$$, if $$\mathcal{A} \vDash \psi [\alpha]$$, then $$\mathcal{A} \vDash \phi [\alpha]$$.

I think I'm right about $$\vDash$$. But I'm less certain about $$\vDash ^*$$. From p. 6 of the book,'$$\mathcal{A}$$ is a model of $$\phi$$' is just:

$$\mathcal{A} \vDash \phi$$

which means

$$\phi$$ is satisfied in $$\mathcal{A}$$ by every assignment.

Therefore, the right hand side of the definition of $$\vDash^*$$

$$\Sigma \vDash ^* \phi$$ iff $$\phi$$ is true in every model of $$\Sigma$$.

is just

For all $$\mathcal{A}$$, for all $$\psi \in \Sigma$$, if $$\mathcal{A}\vDash\psi$$, then $$\mathcal{A} \vDash \phi$$.

However, if what $$\mathcal{A} \vDash \phi$$ means is just that $$\phi$$ is satisfied in $$\mathcal{A}$$ by every assignment, then (and perhaps it is where I have made crucial mistake which I don't understand) I don't see why it cannot be translated also as

For all $$\mathcal{A}$$, for all assignments $$\alpha$$, for all $$\psi \in \Sigma$$, if $$\mathcal{A}\vDash\psi [\alpha]$$, then $$\mathcal{A} \vDash \phi [\alpha]$$.

which is just identical to the definition of $$\vDash$$!

This elementary problem bothers me quite a bit and I hope someone can tell what mistake I have made. Many thanks in advance.

Your analysis of $$\vDash$$ is fine, but you've shuffled some (important) parts of the definition of $$\vDash^*$$. The latter definition says: For every $$\mathcal A$$, if all assignments in $$\mathcal A$$ make all the formulas in $$\Sigma$$ true then all assignments in $$\mathcal A$$ make $$\phi$$ true.

The crucial difference is that here the "all assignments" quantifier is applied separately to the "make all of $$\Sigma$$ true" assumption and the "make $$\phi$$ true" conclusion, whereas in $$\vDash$$ the "all assignments" quantifier is applied to the whole implication "if it makes $$\Sigma$$ true then it makes $$\phi$$ true."

The underlying general fact here is that quantifiers $$\forall x$$ cannot be distributed across implications. $$\forall x\,(P(x)\to Q(x))$$ is not equivalent to $$(\forall x\,P(x))\to(\forall x\,Q(x))$$. Example: It's true that "if all people are American then all people are left-handed" because the antecedent and consequent are both false. But it's not true that "all Americans are left-handed."

• Thanks! I think this is what I want, i.e. the difference in the logical forms of the definitions. Nov 27 '21 at 2:20

I think it's best to take the question's hint and understand that the two notions are not equivalent by looking at a counter-example.

Let $$\Sigma$$ contain only the formula $$\forall x (x + y = x)$$, and let $$A$$ be a model of this formula: i.e., for every assignment of a member of $$A$$'s domain to the variable $$y$$, the formula is true. This is essentially to say that $$x + y = x$$ holds for all $$x$$ and for all $$y$$ in $$A$$, because we can assign any member of the domain to $$y$$. So it turns out that $$A \models \forall y \forall x (x + y = x)$$. This establishes:

$$\Sigma \models ^{*} \forall y \forall x (x + y = x)$$

Is it then also the case that

$$\Sigma \models \forall y \forall x (x + y = x)$$

i.e. for any model $$B$$ and an assignment to $$y$$ from the domain of $$B$$ which satisfies $$\forall x (x + y = x)$$, does $$B$$ also satisfy $$\forall y \forall x (x + y = x)$$? Try to construct a model and assignment where this isn't true. Hint: think of the natural numbers and assign $$0$$ to $$y$$.

The subtle point is that, in reasoning about $$\models ^{*}$$, we require that a model $$A$$ satisfies all of $$\Sigma$$ on any assignment in $$A$$. With plain old $$\models$$, we only require of a model $$B$$ that a particular assignment in $$B$$ satisfies $$\Sigma$$.

• But why then the converse is always true if ⊨ is a lot "looser" than ⊨∗ with open wff(s) in Σ for both cases? Nov 25 '21 at 7:55
• @mohottnad Yep, that was poor wording on my part, which I've removed. Nov 25 '21 at 12:23