Intersection of discriminated union (of records) and a record is incorrect

[RebeccaStevens]

Bug Report

🔎 Search Terms

intersection with discriminated union

This might be a duplicate of #33654 but I'm not completely sure. This issue has a simpler example anyway.
It also seems to be a duplicate of #9919 which was reportedly fixed. This is either regression or a missed case.

🕗 Version & Regression Information

• This is the behavior in every version I tried (v3.3 and v4.5), and I reviewed the FAQ for entries about intersection with discriminated union.

⏯ Playground Link

Playground link with relevant code

💻 Code

type X = ({ a: string } | { b: number }) & { a: unknown };
type Y = X["a"]; // actual: unknown, expected: string

🙁 Actual behavior

type Y is unknown.

🙂 Expected behavior

type Y should be string.

Why:

• When discriminating { a: string } | { b: number } with "a must be defined as a field", we get { a: string }.
• { a: string } & { a: unknown } can be simplified to { a: string }

[RyanCavanaugh]

This is the intended behavior. Through distributivity:

type X = ({ a: string } | { b: number }) & { a: unknown };
// is equivalent to
type X = ({ a: string } & { a: unknown }) | ({ b: number } & { a: unknown });
//                                          ^^^^^^^^^ this one  ^^^^^^^^^^^^

so { b: 3, a: true } is a legal inhabitant of the second constituent of the union, so given an arbitrary value of type X you can't be guaranteed to get a string from it

[RebeccaStevens]

That makes sense.

What's different about the follow case then? How come TypeScript is able to determine that the resulting type is string here?

declare const foo: { a: string } | { b: number };
declare function hasA(value: unknown): value is typeof value & { a: unknown };

if (hasA(foo)) {
    type Z = typeof foo['a']; // string
}

Playground

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.