Allow inferring return type within a type guard in conditional types

Bug Report

🔎 Search Terms

typeguard conditional infer

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about (bug reporting)

⏯ Playground Link

Playground link with relevant code

💻 Code

This is simplified pretty heavily:

type InferredTypeGuard<TypeGuard> = 
    TypeGuard extends ((node: infer Input) => node is infer Output)
        ? (node: Input) => node is Output
        : never;

🙁 Actual behavior

Error on infer Output:

A type predicate's type must be assignable to its parameter's type.
  Type 'Output' is not assignable to type 'Input'.
    'Input' could be instantiated with an arbitrary type which could be unrelated to 'Output'.(2677)

🙂 Expected behavior

TypeScript should allow a conditional type to infer the Output type within the type guard function's return type.

There's no other syntax I can find to do so, and I would have thought this would be the cleanest syntax. Not sure if bug report or feature request...

リンクされた TypeScript Playground の例と、infer Input および infer Output を使用する単純化された条件型から始めます。エラー 2677 を報告する型述語の代入可能性チェックを追跡し、条件型の推論がそれとどのように相互作用すべきかを判断します。例が期待される推論後の入力型と出力型で型チェックを通過し、この動作に対するリグレッションカバレッジが用意されていれば完了です。