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?
Thank you in advance.
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:
$\phi\vDash\bot\iff\forall\mathfrak A[\exists a\in\mathsf{dom}\mathfrak A[\mathfrak A\nvDash\phi[a]]]$