Problem with function types, <T>() => T extends X ? () : ()

🔎 Search Terms

"function", "generic function"

🕗 Version & Regression Information

⏯ Playground Link

https://www.typescriptlang.org/play?#code/C4TwDgpgBAYgdgHgBoD4oF4oICooBQCUGa2UEAHsBHACYDOUSUA-FAIxQBcUATAFChIUAKIBHAK4BDADZ02yADRQAmmnR8osRKjKVq9DVoSqWUYACdx0bgDMZdCHwHhoAdXMB7OAHMAShDpxaWAMEQl7eQBvcwhJGi9pEChJbgByAEFUgF8lSJSoDOyUJydBaHh5HUwcfCJ0El0qWgYmVg5ufjKtHmQ1LFxCYihSCib6RlN23mchMSlZHqQlE3VNCt7G-TpDeB6TVgsrLig7WUcZ6ABhD3MYgGNgf0Dg0Ln7HujY+LhE5LTMnJQPL-IpAA

💻 Code

type Fn<X> = <T>() => T extends X ? 1 : 2
type Equals<X, Y> =
  Fn<X> extends
  Fn<Y> ? true : false

type A = Equals<{readonly a: 'A'}, {a: 'A'}>

🙁 Actual behavior

type A should shows true.

🙂 Expected behavior

type A should be false.

It works well when the code is like this.

type Fn1<X> = <T>() => T extends X ? 1 : 2
type Fn2<X> = <T>() => T extends X ? 1 : 2
type Equals<X, Y> =
  Fn1<X> extends
  Fn2<Y> ? true : false

type A = Equals<{readonly a: 'A'}, {a: 'A'}>

Additional information about the issue

No response

Start with the linked TypeScript Playground and reproduce the minimal conditional generic-function example. Compare the shared Fn alias with the separate Fn1 and Fn2 aliases to isolate why the resulting type differs. Done means type A evaluates to false while the separate-alias example retains its reported behavior.