Suggestion: consolidate generics in union if they match

TypeScript Version: 3.5.2

Search Terms: rxjs startWith generics union consolidate same

Code

declare const create: <T>(
    t: T,
) => {
    startWith: <D>(v1: D) => T | D;
};

// Succeeds
// $ExpectType { foo: number } | { bar: number }
const r1 = create({ foo: 1 }).startWith({ bar: 1 });

// Fails
// $ExpectType { foo: number }
// Actual: { foo: number } | { foo: number }
const r2 = create({ foo: 1 }).startWith({ foo: 1 });

Although the expected/actual types are compatible, it's slightly irritating to deal with a union type T | T instead of a consolidated type T, as it makes type inspection in VS Code much more difficult (especially when the types are large or deeply nested).

Real world use case: https://github.com/ReactiveX/rxjs/issues/4866

Start by reproducing the provided create/startWith example with TypeScript 3.5.2 and inspect the inferred types for r1 and r2. Investigate the type-checking path responsible for unions of identical generic results; done means the matching union is represented as a single type without changing the distinct-union case.