Types that depend on constants are not narrowing correctly

🔎 Search Terms

"ternary narrowing", "ternary constant", "constant narrowing", "constant dependent"

🕗 Version & Regression Information

• This is the behavior in every version I tried (including nightly), and I reviewed the FAQ.

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.2.2#code/MYewdgzgLgBAQgeQQGRgXhlATgVwKYwCGEMokUAUBVAJ4AOBAKgPqO4EYDeMAHgFwxoWAJZgA5gBoYjAUNFiYAX2r0mzAGKEANhA4xu-QdnlT1s4+KVVaDaehgAKG3hAAzeEmQBKGHh5Q8MAATEmx8GAB+aVZ2GAEWTR08KgpXHDBgKGFwGFcHL3j9ChgS0nBoGAA3QiwZaLZwrl4BAHJq2papOpaxEBAglqtSsvIqmrNoxN17A1b29U6YCZaQAGtCGkHlYtKsPCgcLDAPFEix2rjz9QpFIA

💻 Code

const BOOL = true as const

type T_True = { x: string, T: string }
type T_False = { x: string, F: string }

type T = (typeof BOOL) extends true ? T_True : T_False

function f(): T {
    const varT: T_True = { x: 'varT', T: 'good' }
    const varF: T_False = { x: 'varF', F: 'okay' }

    return BOOL ? varT : varF
}

🙁 Actual behavior

Typechecking fails with the error Type 'T_True | T_False' is not assignable to type 'T_True'. Property 'T' is missing in type 'T_False' but required in type 'T_True'. on the return statement.

🙂 Expected behavior

Typechecking should pass, since the const BOOL = true as const should mark the false ternary case as unreachable.

I also tried making T generic with respect to BOOL, and splitting the ternary operator into an if/else, but the same error persisted regardless.

Additional information about the issue

No response

Start by running the linked TypeScript Playground with the supplied BOOL, conditional type, and ternary example to reproduce the return-statement error. Trace the type-checking behavior for constant conditions; done means the example type-checks with the false branch treated as unreachable.