`extends any ?` may return right hand in conditional types

TypeScript Version: 4.0.5

Search Terms:

• type inference
• extends any
• conditional types
• right hand

Expected behavior:

type D1 = X1 <O1> returns false.

Actual behavior:

In the code below, type D1 = X1<O1> seems to return false, but it actually returns true.
If you change the seemingly irrelevant right-hand type in extends any ? written for union distribution to undefined, it will behave as intended.

Related Issues:

Code

type X1<T> = IfEq<
  AllKeys1<T>,
  keyof T,
  true,
  false
>;

type X2<T> = IfEq<
  AllKeys2<T>,
  keyof T,
  true,
  false
>;

type IfEq<X, Y, T, E> = [X] extends [Y] ? ([Y] extends [X] ? T : E) : E;

// AllKeys<{ x: ... } | { x: ...; y: ... }> -> "x" | ("x" | "y") -> "x" | "y"
type AllKeys1<O extends Object> = O extends any ? keyof O : never;
type AllKeys2<O extends Object> = O extends any ? keyof O : undefined;

type O1 = { x: string; y: number } | { x: boolean; z: number }
type A1 = AllKeys1<O1> // "x" | "y" | "z"
type A2 = AllKeys2<O1> // "x" | "y" | "z"
type B = keyof O1 // "x"

type C1 = IfEq<A1, B, true, false> // false
type C2 = IfEq<A2, B, true, false> // false
type D1 = X1<O1> // true ???
type D2 = X2<O1> // false

type O2 = { x: string; y: number } | { x: boolean; y: Object }
type E1 = X1<O2> // true
type E2 = X2<O2> // true

"use strict";

{
  "compilerOptions": {
    "noImplicitAny": true,
    "strictNullChecks": true,
    "strictFunctionTypes": true,
    "strictPropertyInitialization": true,
    "strictBindCallApply": true,
    "noImplicitThis": true,
    "noImplicitReturns": true,
    "alwaysStrict": true,
    "esModuleInterop": true,
    "declaration": true,
    "experimentalDecorators": true,
    "emitDecoratorMetadata": true,
    "moduleResolution": 2,
    "target": "ES2017",
    "jsx": "React",
    "module": "ESNext"
  }
}

Playground Link: Provided

TypeScript 4.0.5 を使用して、提供された Playground の再現から始め、D1、D2、E1、E2 の条件型の結果を比較します。コンパイラーによる分配条件型の処理を追跡し、報告された期待される結果と実際の結果が、他のケースを変更せずに一貫して解決されることを確認します。