Prove that theory is not Henkin one

Question

The definition as it was given to me: The theory $T$ is Henkin theory, if and only if for every formula $\phi$ in $T$ we have constant $c$ language of $T$ such as $T \vdash \exists x \phi \to \phi(x/c)$. How I have language $\mathcal L = \{ c \}$, where $c$ is a constant, and I have to prove that theory $T = \{ \emptyset \}$ over language $\mathcal L$ is not a Henkin one. My problem is that I don't know how to even prove this, if there is no formulas in the language to "check Henkin condition". How do I approach this task?

Answer 1

If the theory $T$ has no non-logical axioms (this is my reading of $T=\{ ∅ \}$) then - by soundness - it can prove only valid formulae.

Assuming that $T$ is an Henkin theory, then for every formula $ϕ$ in the language of $T$ we have a constant $c$ such that $T ⊢ ∃xϕ→ϕ[x/c]$.

Thus, also : $T \vdash ∃x \ \lnot \phi \to \lnot \phi [x/c]$, and then, by contraposition :

$T \vdash \phi [x/c] \to \forall x \ \phi$

which is clearly invalid.