# Does $\phi\vDash\bot$ imply that $\vDash\phi\to\bot$ if $\phi$ is a formula that has free variables?

This question is a sequel of this one where I asked the same thing with $$\vDash$$ replaced by $$\vdash$$.

Inspired by comments received on that question I switched from syntax to semantics.

Let $$\mathcal{L}$$ be a first order language.

Let $$\phi$$ denote a $$\mathcal L$$-formula that has at least one free variables.

Purely for convenience let us only look at the case where it has exactly one free variable $$x$$.

If my understanding is okay then:

• $$\phi\vDash\bot$$ iff every $$\mathcal L$$-structure $$\mathfrak{A}$$ has some element $$a$$ in its domain such that $$\phi\left[a\right]$$ is false in $$\mathfrak{A}$$. This because only in that situation no $$\mathcal L$$-structure $$\mathfrak A$$ exists that satisfies $$\mathfrak A\vDash\phi$$.

• $$\vDash\phi\to\bot$$ iff for every $$\mathcal L$$-structure $$\mathfrak{A}$$ and every element $$a$$ in its domain statement $$\phi\left[a\right]$$ is false in $$\mathfrak{A}$$. This because only in that situation $$\mathfrak A\vDash\phi\to\bot$$ for every $$\mathcal L$$-structure $$\mathfrak A$$.

Unfortunately it is not clear that $$\phi\vDash\bot$$ implies that $$\vDash\phi\to\bot$$ and I even wonder whether that's true.

Could you set straight wrong understandings or take away a blind spot (if there is any) please?

Addendum to make clear where my understanding of $$\phi\vDash\bot$$ comes from.

• $$\mathfrak A\vDash\phi\iff\forall a\in\mathsf{dom}\mathfrak A[\mathfrak A\vDash\phi[a]]$$ (1.7.9 Leary)
• $$\phi\vDash\psi\iff\forall\mathfrak A[\mathfrak A\vDash\phi\implies\mathfrak A\vDash\psi]$$ (1.9.1 Leary)

Taking $$\bot$$ for $$\psi$$ in the last bullet we get:

$$\phi\vDash\bot\iff\forall\mathfrak A[\mathfrak A\nvDash\phi]$$

Then applying the first bullet we arrive at:

$$\phi\vDash\bot\iff\forall\mathfrak A[\exists a\in\mathsf{dom}\mathfrak A[\mathfrak A\nvDash\phi[a]]]$$

We do indeed have $$\phi\models\perp\quad\iff\quad\models\phi\rightarrow\perp.$$

Your understanding of $$\phi\models\perp$$ is incorrect: we have $$\phi\models\perp$$ iff for every structure $$\mathcal{M}$$, every variable assignment which makes $$\phi$$ true makes $$\perp$$ true. Since no assignment can make $$\perp$$ true, this means that there is no structure $$\mathcal{M}$$ and variable assignment making $$\phi$$ true - or in other words, no structure has any tuple satisfying $$\phi$$.

And this clearly matches up with $$\models\phi\rightarrow\perp$$ (your analysis of this is correct).

EDIT: Specifically, the issue is that your definition of $$\phi\models\psi$$ in the variables-allowed context is incorrect: the "quantification over valuations" has to happen outside the $$\models$$-part on the right hand side.

The right definition is $$\forall \mathfrak{A}, a\in\mathfrak{A}(\mathfrak{A}\models\phi[a]\implies\mathfrak{A}\models\psi[a]).$$ On the other hand, the relation you've defined - which I'll call "$$\models_?$$" for clarity - is equivalent to the following: $$\forall\mathfrak{A}[\forall a\in\mathfrak{A}(\mathfrak{A}\models\phi[a])\implies \forall a\in\mathfrak{A}(\mathfrak{A}\models\psi[a])].$$ To see the difference between these, consider the following formula in the language consisting of a single unary relation symbol $$U$$:

$$\phi(x):\quad$$ If $$U$$ describes a nonempty proper subset of the domain, then $$U(x)$$.

You can check that we have $$\phi(x)\models_?\phi(y)$$, which clearly should not hold.

And this accounts for the apparent discrepancy in the OP. Using the right definition, we have $$\phi\models\perp$$ iff $$\forall \mathfrak{A},a\in\mathfrak{A}(\mathfrak{A}\models\phi[a]\implies \mathfrak{A}\models\perp)$$ iff $$\forall \mathfrak{A}\color{red}{\forall} a\in\mathfrak{A}(\neg\mathfrak{A}\models\phi[a])$$ as desired.

• I have added the reasonings that led me to this (mis)understanding. Would you be so kind to take a look at them, Noah? They are based on what I encountered in "Leary and Kristiansen". I am still puzzled. – drhab Aug 24 '20 at 19:05
• @drhab See my edit (your definition of $\models$ for formulas has a quantifier-placement error). – Noah Schweber Aug 24 '20 at 19:24
• Thank you very much. One of the coming days (now it is time to get some sleep) I will probably add a citation from "Leary and Kristiansen" in order to get things straight. I will let you know. I do understand the difference between $\vDash$ and $\vDash_{?}$ but am still a bit shaky. Good night. – drhab Aug 24 '20 at 19:48
• In Leary it is stated that for applying the deduction theorem we must be dealing with a sentence. That seems redundant under your definition. Further a rule of interference is there: $(\{\psi\to\phi\},\psi\to\forall x\phi)$ under condition that $x$ is not free in $\psi$. The soundness of it can be proved with $\vDash_{?}$ but under your $\vDash$ the rule seems not to be sound. Also there is an exercise of the form: "prove that $\phi\vDash\psi$ but not $\vDash\phi\to\psi$. – drhab Aug 25 '20 at 8:37
• What I said in my former comment about deduction theorem is a matter of syntax, so might not be relevant here. But the fact remains that your definition does not seem to match with the one practicized by Leary (nor Mendelsohn). – drhab Aug 25 '20 at 10:08