MW: An addendum to the last post. I do have an employment opportunity for one of those pathological scaffolds: the one where B(0) is the 2-element boolean algebra, and all the B(n)’s with n>0 are trivial. It’s perfect for the semantics of a structure with an empty domain.
The empty structure has a vexed history in model theory. Traditionally, authors excluded it from the get-go, but more recently some have rescued it from the outer darkness. (Two data points: Hodges’ A Shorter Model Theory allows it, but Marker’s Model Theory: An Introduction forbids it.)
All n-ary predicates with n>0 are universally true, because they’re vacuously true. But ∃x(x=x) is false, while ∀x(x=x) is true. So B(0) contains both verum ⊤ and falsum ⊥.
The empty structure looks a little sketchy in this sense: the standard rules of logic require modification for it. (That’s why you were able to prove that its logic isn’t a hyperdoctrine.) On the other hand, its good friends the number 0 and the empty set will vouch for its good character. Hodges allows it because it makes some theorems nicer.
JB: This is very interesting. Near the end of our last conversation I claimed without proof that the Frobenius condition implies
∃x(⊤) = ⊤.
I now think this was wrong. Without this, I don’t think we can rule out the hyperdoctrine where B(0) is the 2-element boolean algebra but all the B(n)’s with n>0 are trivial. And that makes me happy, because I really don’t want to rule out the empty model of first-order theories—that is, what you’re calling the ’empty structure’.
So let me revise some of my statements from the last episode. I want to define an inconsistent hyperdoctine B to be one with ⊤ = ⊥ in B(0), or equivalently, one where B(0) is the trivial boolean algebra. This implies that all the B(n)’s with n>0 are trivial.
If any B(n)’s with n>0 is trivial, then they all are—our techniques from last time do show this—but this does not imply that B(0) is trivial. There is a hyperdoctrine where B(0) is the 2-element boolean algebra but all the B(n)’s with n>0 are trivial. And this one is consistent.
To really prove this we should dig deeper into the Frobenius condition—and for that matter, the Beck-Chevalley condition. If I try to explain these to you, that will force me to understand me better.
But I think we’re okay for now. So, this time it’s for real: on to Gödel’s completeness theorem!
Diagonal Argument 
I’m so sad this didn’t continue. You could definitely pick it up from where you left it. Thanks for posting up to this point anyway, this was very helpful and illuminating.