为什么在这种情况下总是true条件类型返回不进行类型检查

我发现

type B = { 
  foo: string;
  bar: number;
};

function get<F extends B, K extends keyof B>(f: F, k: K): F[K] {
  return f[k];
}

直觉上与

function get<F extends B, K extends keyof B>(f: F, k: K): F extends B ? F[K] : undefined {
  return f[k];
}

但后者不进行类型检查:

Type 'F[K]' is not assignable to type 'F extends B ? F[K] : undefined'.
  Type 'F["foo"] | F["bar"]' is not assignable to type 'F extends B ? F[K] : undefined'.
    Type 'F["foo"]' is not assignable to type 'F extends B ? F[K] : undefined'.
      Type 'string' is not assignable to type 'F extends B ? F[K] : undefined'.

因此有一个问题:TS编译器对我不明显的类型了解多少?

游乐场: http://www.typescriptlang.org/play/#code/C4TwDgpgBAQlC8UDeUBQUoDMD22BcUAzsAE4CWAdgOYDc6UARgIYkEUCuAtgxCXQL51M7CgGNgZbBShUIwADwAxKBAAewCBQAmhWABooAaRXrNOqAGsIIbJlgA+ABSYCigxYKGAlK5MbtunAA-FCKANqGALpQBCJaEJiUEFrI9CRy7CTSmGEWkQKoQA