The type of inference is inaccurate about typescript

[Andarist]

Inference is based on a set of heuristics. Currently, TS eliminates matching types here (based on isTypeIdenticalTo) and when either targets or sources are emptied... it returns. Later on, it picks up the constraint of your type (unknown) variable in the absence of inference candidates.

This is a "correct answer" since unknown & string is still string

[Zackin-He]

For example, in the following two examples, I still can't know why the result of the first example is unknown instead of string, and why I can't get a more precise type like the second example.
Where did the targets or sources and inference candidates you mentioned come from.

type MyType<T> = T extends (infer R) & string ? R : never
type Result = MyType<string> // unknown

type MyType<T> = T extends (infer R) & string ? R : never
type Result = MyType<'abc'> // abc

[Andarist]

It's just how TypeScript works today. It has some algorithm for matching between your source (type argument here) and the targets (the intersection here). There is no better answer to that - it's an implementation detail. It's not incorrect though.

Those examples aren't really practical, so I'm not sure what you're trying to solve here. I understand that it might be confusing and interesting why they behave differently but that's about it. I'm sure that this part of the algorithm was written to improve some more real-life scenarios for which that selection mechanism makes sense more often than not.