Second-order strong minimality and amorphousness, take 2 Recently I asked a question about whether a second-order analogue of strong minimality could correspond to amorphous satisfiability (= having a model whose underlying set cannot be partitioned into two infinite pieces). I missed a fairly obvious fact: roughly speaking, that in a second-order way we can assert the existence of a certain type of structure on the domain prohibiting amorphousness (e.g. a linear order or a non-surjective endofunction) without making any such structure definable.
I'd like to rephrase my original question to avoid this. One idea which seems promising is to demand something like a second-order witness property: whenever our theory asserts the existence of some structure on the domain there should be some definable such structure. The existence of a second-order sentence saying "The domain is amorphous" causes a bit of difficulty in making the question nontrivial, however (e.g. any maximally satisfiable second-order theory either proves that the domain is amorphous, in which case it obviously is amorphously satisfiable, or proves that there is a partition of the domain into two infinite piecs at which point the second-order witness property kills any type of minimality).
The easiest way to get around this, in my opinion, is to restrict attention to monadic theories. My question is whether it is consistent with $\mathsf{ZF}$ that every satisfiable monadic second-order theory $T$ with the following properties is amorphously satisfiable:

*

*$T$ is negation-complete: for each $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$.


*$T$ is "second-order strongly minimal:" no model of $T$ has a definable-with-(element-)parameters bi-infinite subset.


*$T$ has the "second-order witness property:" whenever $\exists X\varphi\in T$, there is some unary predicate symbol $U$ such that $\varphi[X/U]\in T$.
Note that something like the witness property is necessary here, per Harry West's answer to my original question.
 A: I claim that if the language is countable, or merely well-orderable, then the witness property by itself prevents amorphous domains, and even infinite Dedekind-finite domains. Strong minimality has nothing to do with it.
Theorem. Any satisfiable complete theory $T$ in a well-ordered language with the witness property is never true on an amorphous domain, nor even on an infinite Dedekind-finite domain.
Proof. Suppose the language is well-ordered and complete theory $T$ has the witness property and is true in some infinite domain $M$. Since
$$\exists X(X\text{ has exactly one element })$$ is true, then by the witness property there is a predicate $U_1$ whose extension is a singleton. Now,
$$\exists Y(Y\text{ has exactly two elements, one of them in }U_1)$$
is true, so there is a predicate $U_2$ having exactly one additional element beyond the element of $U_1$. And further,
$$\exists Z(Z\text{ has exactly three elements, two of them in }U_2)$$
is true, so there is a tripleton set $U_3$. And so on.
In this way, we construct a countably infinite subset of the domain. So the domain is not Dedekind finite. $\Box$
We used the well-orderability of the language when picking the particular predicates $U_k$, since there could be many witnessing predicates for the witness property.
The theorem shows that in order to have a theory with the witness property true on an amorphous domain, the language itself will have to be a little strange, perhaps even containing an amorphous set of unary predicate symbols.
