If you're not careful about how you encode a property, the type system will _require_ you to preserve that property throughout each step of the computation, reg

If you're not careful about how you encode a property, the type system will _require_ you to preserve that property throughout each step of the computation, regardless of what sort of morphism that computation is. (It's an endomorphism.) Now that may be simple and straightforward when dealing with structural recursion over a boolean language, but if you want to write an inliner for Haskell, for example, it's probably a better idea to drop invariants from types (but not from the computation, of course), run the algorithm, and reintroduce invariants to the types afterwards. Again, that example's way overshooting the complexity needed to prove the point. Even a seemingly simple invariant like the red-black balancing property is one that is impractical to preserve in this fashion without a proof assistant. And even then its practicality is still dubious.