Type predicate assignable to the constraint of type 'A', but 'A' could be instantiated with a different subtype of constraint '{}'

Question

I have a type that takes an object and marks one of its properties as being unique:

type UniqueString = string & { readonly UniqueString: unique symbol }

type HasUnique<A extends {}, K extends keyof A> = Omit<A, K> & Record<K, UniqueString>

I'm trying to write a type guard to take an A and return is HasUnique<A, K>. Here is my type guard, which gives me a type error:

declare const hasUnique: <A extends {}, K extends keyof A>(a: A) => a is HasUnique<A, K>

TS2677: A type predicate's type must be assignable to its parameter's type. Type 'HasUnique<A, K>' is not assignable to type 'A'. 'HasUnique<A, K>' is assignable to the constraint of type 'A', but 'A' could be instantiated with a different subtype of constraint '{}'.

Is there a way to fix this, or is this one of those instances where I've hit the limitations of TS's inference with generics, as referenced in this SO question?

It is not a limitation. It just how custom typeguards work. How do you call this function? There is unused `K` generic. Do you call this function with explicit two generics? — captain-yossarian from Ukraine, Jan 18 '22 at 19:54
`Omit` is a widening operation and type guards must be narrowing. If I have `declare const obj: { a: "hello" & { readonly UniqueString: unique symbol } };` and call `hasUnique(obj)` then a `true` result would presumably be *widening* the `a` property to forget that it has a literal string type. Why not just `A & Record` instead of `Omit & UniqueString`? What's the use case? Can you show these uses in a [mre]? — jcalz, Jan 18 '22 at 20:46

Type predicate assignable to the constraint of type 'A', but 'A' could be instantiated with a different subtype of constraint '{}'

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