incorrect inference in union of tuple types

🔎 Search Terms

tuple type union, inference, discriminator

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about Generics.

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.3.3#code/C4TwDgpgBAggdgEwDwBUB8UC8UDaAiAQ0TwBooUIAnAW1TTIproF0AoUSKAUQEc6tceCD1Lk2HaAElgAZ1RkA0lAgAPYBEQyoAawggA9gDNyGbPgCWs0QoZVaKHAuZpx4aI3unufdFAA+UNJyKGS6Bsa+AfDI6KyshgCucADGwOb6cFBglBAI5skE6nQAFOo0AFzkdnQAlFAA3qxQzVA5wAmUmWXUrAC+cckZMsBQwJUeSPUE5XAJ1ABGVL1e2bn5hRDFFlZkhKL4wqIADMzMNaxAA

💻 Code

type And<T> = ["and", Term<T>, Term<T>]
type Eq<T> = ["eq", T]
type Its<T, K extends keyof T> = ["its", K, Term<T[K]>]
type Term<T> = Eq<T> | Its<T, keyof T> | And<T>

function predicate<T>(term: Term<T>) {
    return term
}

const t: Term<{a:number}> = predicate(["its", "a", ["eq", 0]])

🙁 Actual behavior

TypeScript complains that Type "a" is not assignable to type number | "toString" | "toFixed" | "toExponential" | "toPrecision" | "valueOf" | "toLocaleString" | Term<number>. The expected types are a union of number from Eq<number>, Term<T> from And<T> and keyof number from Its<number>. Eq<number> can be ruled out by both length and first element and And<T> by the first element. number was chosen as value of T despite there being a solution that is also clear for the expected return type.

🙂 Expected behavior

TypeScript infers T to be { a: number } (or { a: any } when the return type is not known).

Additional information about the issue

It works when you remove And from Term, moving the "and" to the last index or when modelling something similar using objects instead of tuples. The first element could be used to discriminate the union.

Reproduziere den Tupel-Union-Inferenzfehler im verknüpften TypeScript Playground unter Verwendung der bereitgestellten Term-, Eq-, Its- und And-Typen. Untersuche, wie das erste Tupel-Element beim Inferieren des generischen Parameters über die Union hinweg verwendet wird. Als abgeschlossen gilt die Aufgabe, wenn das Beispiel T als { a: number } (oder als { a: any } ohne einen kontextuellen Rückgabetyp) inferiert und dabei die bestehenden Tupel-Fälle erhalten bleiben.