Nominal/branded key for `Record` disables checking/inferring of value type

Bug Report

🔎 Search Terms

nominal, Record, branded

🕗 Version & Regression Information

3.3.3 through 4.3.2

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

⏯ Playground Link

Playground link with relevant code

💻 Code

type Id = string & { __id: 'id' }

type User = {
  id: Id;
  name: string;
}

type Users = Record<Id, User>;

type Users2 = Map<Id, User>;

type Users3 = Record<string, User>;

// This will not produce an error
// @ts-expect-error No name should be an error!
const users: Users = { id0: { id: 'id', } }

// type: [string, unknown][]
const entries = Object.entries(users);

const users2: Users2 = new Map();
// @ts-expect-error No name produces an error
users2.set('id' as Id, { id: 'id' as Id, })

// @ts-expect-error No name produces an error
const users3: Users3 = { id0: { id: 'id' as Id, } }

// type: [string, User][]
const entries3 = Object.entries(users3);

🙁 Actual behavior

Using a nominal/branded type as the key to a record type causes type checking of the value to be skipped. Additionally, inference will also fail and fallback to unknown, such as using Object.entries. Using a Map works as expected, but not a Record. I realize that the runtime behavior between a Map (real) and Record (type) is completely different, I'm just providing it as an example.

🙂 Expected behavior

I'd expect types to be preserved and checked, even when using a nominal/branded type as a key to a record.

Empieza ejecutando la reproducción enlazada de TypeScript Playground con los ejemplos Record<Id, User> y Object.entries, y compáralos después con los casos funcionales Map y Record con claves de tipo string. Sigue el procesamiento que hace el compilador de las claves branded y de los valores de los records; el trabajo estará terminado cuando la comprobación de valores informe de la falta de name y Object.entries(users) conserve User en lugar de inferir unknown.