Discriminated union containing optional properties and generic types is not narrowed

Bug Report

I would have expected doesNotNarrow (below) to also be able to narrow x.b to NonUndefined<B>.

🔎 Search Terms

generic, narrow, discriminated, union, optional

🕗 Version & Regression Information

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

⏯ Playground Link

Playground link with relevant code

💻 Code

 type NonUndefined<T> = T extends undefined ? never : T

type X<A, B> = { a: NonUndefined<A>, b?: never } | { a?: never, b: NonUndefined<B> }

function doesNarrow(x: X<number, number>) {
  if (x.a === undefined) {
    const y = x.b
      // ^? number
  } 
}

function doesNotNarrow<A extends number, B extends number>(x: X<A, B>) {
  if (x.a === undefined) {
    const y = x.b
      // ^? NonUndefined<B> | undefined
  } 
}

🙁 Actual behavior

x.b was not narrowed.

🙂 Expected behavior

I expected x.b to be narrowed because if x.a is undefined, then x must be the latter part of the discriminated union because x.a can only be undefined there. TypeScript seems to be able to work that out when the type parameters are constant/non-generic (see doesNarrow), but for some reason it's not able to work it out for the generic case (even though NonUndefined should make it feasible).

Start with the linked TypeScript Playground and compare the narrowing in doesNarrow with doesNotNarrow. Investigate the generic discriminated-union narrowing path using the provided X, NonUndefined, and optional-property example. Done means x.b is narrowed to NonUndefined in the generic case without regressing the non-generic behavior.