Inferred type symbol in conditional type should be visible-but-banned in falsy arm

[RyanCavanaugh]

Currently if you write

type C<T> = T extends infer U ? U : U;
//                                  ~ Cannot find name "U"

We give you an error because U is not in scope at all in the false arm.

This is a) confusing (U is right there!) and b) a landmine, because you might have an outer type of the same name

// Elsewhere
type U = string;
// ...
/// Not what I meant!
type C<T> = T extends infer U ? number : U;

We should keep U in scope in both cases and disallow its use with an explicit error message, e.g. (please bikeshed):

Inferred type parameter "U" is unmatched in this case and cannot be used

This should not block nesting of lexical scopes; i.e. this should still be legal because each U would shadow the outer expression's:

type C<T> = T extends Array<infer U> ? U :
            T extends Promise<infer U> ? U :
            never;

[RyanCavanaugh]

cc @ahejlsberg @DanielRosenwasser since we should 👍 / 👎 before release

[krryan]

Why keep U in scope in the falsy case at all? Why not allow the use of the outer U? Just for readability/least-surprise concerns? I can buy that but I'm wondering if I'm missing something more.

[RyanCavanaugh]

This one missed the boat