The title question is too vague so let me be specific.


Much of modern finite semigroup theory uses profinite semigroups and properties of profinite semigroups that depend on the existence of prime ideals in Boolean algebras, which is a choice principle weaker than AC. Often results are proved first via profinite methods and then explicit proofs using finite means are found later. 


To narrow my question down, let me consider the following situation. Let V be what is sometimes called a variety of finite semigroups, that is, a class of finite semigroups closed under finite products, subsemigroups and homomorphic images. A pro-V semigroup is an inverse limit of semigroups in V. It is known, using Tychnoff's theorem for products of finite spaces, that every continuous finite quotient of a pro-V semigroup belongs to V. So if you want to show a variety W of finite semigroups is contained in a variety V it suffices to show each element of W is a continuous quotient of a free pro-V semigroup. Sometimes one has good information on the structure of free pro-V semigroups and can exploit this. 


For example, Almeida proved that the smallest variety of finite semigroups containing all finite commutative semigroups and all finite groups is the variety of finite semigroups with central idempotents using the approach sketched above. Later Auinger gave a proof using only finite semigroups. 



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> > **Question.** Is there some general result in logic or set theory that would imply that the existence of a proof in ZF+Boolean prime ideal theoerem that a variety W of finite semigroups is contained in a variety V of finite semigroups implies the existence of a proof in ZF that W is contained in V?
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Many results of this sort in finite semigroup theory were motivated by questions in automata theory and in principle one would like to avoid choice in this context. 


**Caveat:** I know very little set theory or model theory so please take that into account in answer.



Is there a general result that theorems about finite structures proved in ZFC can be proved in ZF?
As long as your varieties have reasonable definitions, the axiom of choice will be eliminable from proofs like these. The point is that any finite semigroup has an isomorphic copy whose underlying set consists of natural numbers, and that copy will be in Gödel's constructible universe $L$, where the axiom of choice holds. So if there were a semigroup in $W$ that isn't in $V$, then the same would be true in $L$.


Now about that proviso "reasonable definitions": You can sneak a lot of set theory into the definition of a variety. Consider the variety that consists of all groups if the continuum hypothesis holds and consists of all commutative semigroups if the continuum hypothesis fails. With such "cheating" you can surely arrange for $W$ to be included in $V$ iff the axiom of choice holds. My argument in the preceding paragraph tacitly assumed that the definitions of the varieties were absolute, in the sense that the constructible universe $L$ and the whole universe agree as to whether any particular finite semigroup is in the variety. By working harder, one can get by with weaker absoluteness requirements, but one can't get rid of them completely.


