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Conditional types behavior is different when referencing same type with different nameΒ #44945

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@k8w

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@k8w

Bug Report

πŸ”Ž Search Terms

conditional types different name

πŸ•— Version & Regression Information

v4.4.0-beta

  • This is the behavior in every version I tried, and I reviewed the FAQ for entries about conditional types

⏯ Playground Link

Playground link with relevant code

πŸ’» Code

type Left<Z> = <T>() => (T extends Z ? 1 : 2);
type Right<Z> = <T>() => (T extends Z ? 1 : 2);

type Equal1<X, Y> = Left<X> extends Right<Y> ? true : false;
type Equal2<X, Y> = Left<X> extends Left<Y> ? true : false;

// false
type X1 = Equal1<{a:string}, {readonly a: string}>
// true
type X2 = Equal2<{a:string}, {readonly a: string}>

πŸ™ Actual behavior

X1 and X2 is different.

πŸ™‚ Expected behavior

X1 and X2 is the same.

Actually, Left and Right are the same, Equal1 and Equal2 are the same.
But the result of Equal1 and Equal2 with the same parameters are different.

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BugA bug in TypeScriptFix AvailableA PR has been opened for this issueRescheduledThis issue was previously scheduled to an earlier milestone

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