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...

Comienza con el ejemplo vinculado de TypeScript Playground y el tipo condicional simplificado que usa infer Input y infer Output. Traza la comprobación de asignabilidad del predicado de tipo que informa del error 2677 y determina cómo debería interactuar con ella la inferencia de tipos condicionales. Se considera terminado cuando el ejemplo se comprueba correctamente con los tipos de entrada y salida inferidos esperados, y existe cobertura de regresión para este comportamiento.