Make identity comparison reflexive across unions or intersections of the same type

[MichaelMitchell-at]

Please verify that:

• There is an associated issue in the Backlog milestone (required) - No one has created an issue for this specifically, but it is a subset of a larger set of problems discussed in [Feature request]type level equal operator #27024 and can be seen as complementary to Merge trivially mergeable intersection types for identity comparison #60726
• Code is up-to-date with the main branch
• You've successfully run hereby runtests locally
• There are new or updated unit tests validating the change

#60726 (comment) points out why some type testing libraries define equality with a definition like:

type Equals<A, B> = (<T>() => T extends (B & T) | T ? 1 : 0) extends (<T>() => T extends (A & T) | T ? 1 : 0) ? true : false;

rather than just

type Equals<A, B> = (<T>() => T extends B ? 1 : 0) extends (<T>() => T extends A ? 1 : 0) ? true : false;

which is that types like {a: 1} are not considered identical to types like {a: 1} & {a: 1} and {a: 1} | {a: 1}.

The latter definition of Equals is required to enable #60726, so I've made this PR to eliminate the need for the former definition of Equals.

[typescript-bot]

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