How to compose union type in a generic function?

Question

Let's have the following simplified example:

type AB = {
    foo: "A"
    bar: "A"
} | {
    foo: "B"
    bar: "B"
}

type Foo = AB["foo"]

function fun<F extends Foo>(foo: F, bar: Extract<AB, { foo: F }>["bar"]) {
    fun2({
        foo,
        bar,
    })
}

function fun2(ab: AB) {

}

It fails with: Argument of type '{ foo: "A" | "B"; bar: "A" | "B"; }' is not assignable to parameter of type 'AB'.

TS playground

1. How to restrict the generic type not to accept union types?

   • @jcalz answers a similar question here Constraining type in Typescript generic to be one of several types. But the answer requires passing all "simple types" to a tuple type.
   • @jcalz then discourages to use his solution from https://stackoverflow.com/a/55128956/121968 to convert union type to tuple type.

2. And then how to prevent bar: Extract<AB, { foo: F }>["bar"] to return union type?

   • fun<F extends "A"> will fail with Argument of type '{ foo: F; bar: "A" | "B"; }' is not assignable to parameter of type 'AB'.

     (Btw. why foo: F and not just foo: "A" ?)

   • But bar: Extract<AB, { foo: "A" }>["bar"] will work.