Why is this conditional based on a non-distributive union type not working?

I have an interface and want to determine if any of the values are optional. To do that, I thought I'd create a union out of all the values and then check if it extends undefined. When it didn't work, I read about distributive conditionals and thought surely that would be the solution.

Here's the code:

export interface SomeInterface {
  someBool: boolean
  someOptionalNum?: number
  someOptionalText?: string
}

const test: [SomeInterface[keyof SomeInterface]] extends [undefined] ? boolean : string = true

According to this part of the typescript guide, I would expect this code to work. But I can't get test to ever be a boolean, it remains stuck on string.

See Typescript Playground

Our understanding of the way extends works is causing the issue here.

Conditional types take a form that looks a little like conditional expressions (condition ? trueExpression : falseExpression) in JavaScript:

  SomeType extends OtherType ? TrueType : FalseType;

When the type on the left of the extends is assignable to the one on the right, then you’ll get the type in the first branch (the “true” branch); otherwise you’ll get the type in the latter branch (the “false” branch).

Take a look at this code for example:

type X = "A" | "B";
type A = "A";
type Y = A extends X ? true : false;

Y would be true above.

In the question, [SomeInterface[keyof SomeInterface]] is not assignable to [undefined]. In fact, [undefined] is assignable to [SomeInterface[keyof SomeInterface]].

So you have the conditions the other way round. You can change your code like:

export interface SomeInterface {
  someBool: boolean
  someOptionalNum?: number
  someOptionalText?: string
}

const test: [undefined] extends [SomeInterface[keyof SomeInterface]]  ? boolean : string = true

and it should work.

Playground Link