Why `T extends "A"` and `S extends "A"` have no overlap?

Issue

I'm confused with a typescript compile error abut the following code:

function f<T extends "A", S extends "A">(x: T, y: S) {
  if (x === y) {
     // ERROR: This condition will always return 'false' since the types 'T' and 'S' have no overlap. ts(2367)
  }
}

playground

Obviously, T and S could have overlap, but the compiler says they have no overlap. Why does this happen and how to resolve this error?

Solution

I guess this comment by RyanCavanaugh also applies to your case:

=== is only allowed between two operands if there is a "comparable" type relationship between them. This is similar to the check that determines if a type assertion is legal, with some extra special cases.

To see a more meaningful error from the compiler, try a cast:

function f<T extends "A", S extends "A">(x: T, y: S) {
  /**
   * Conversion of type 'T' to type 'S' may be a mistake 
   * because neither type sufficiently overlaps with the other. 
   * If this was intentional, convert the expression to 'unknown' first.
   * 
   * 'T' is assignable to the constraint of type 'S',
   * but 'S' could be instantiated with a different subtype of constraint '"A"'.
   */  
  const a = x as S;
}

To make the compiler happy, you can convert to unknown or any.

  if (x as unknown === y) {
    return true;
  }

Playground

Answered By - TmTron