This is perhaps a stupid question, but I ask it anyway. It seems to me that the semantic comes after and it cannot change the complexity of the language. I ask the question, because Herbert B. Enderton in https://plato.stanford.edu/archives/spr2008/entries/logic-higher-order/ wrote:
We have seen that, although the set V¹ of valid formulas of first-order logic is computably enumerable, the corresponding set V² for second-order logic (with the standard semantics) is vastly more complex. This phenomenon does not continue into the higher orders.
Why did he had to specify "with the standard semantics"? I am learning and checking my understanding, perhaps on a detail. The context is that, with the Henkin semantic, this language is equivalent to many-sorted first order logic with the comprehension axioms. Is it correct to say that, nevertheless, the language itself remains non recursively enumerable, even if it is equivalent (in terms of interpretations) to a language that is recursively enumerable? I am checking if I have missed something.