Variadic element of tuple-like intersection types are spreading incorrectly

TypeScript Version: 4.0.3

Search Terms: variadic tuple, variadic element, intersection type, tuple-like type

Code

type T1 = [...unknown[] & []]; // never[]
type T2 = [...number[] & [1, 2, 3]]; // (3 | 1 | 2)[]

Expected behavior: for T1, I expected it to be [](empty tuple type) and for T2, I expected it to be [1, 2, 3].

Actual behavior: Both T1 and T2 became Array rather than tuple somehow. (T1 is deduced to never[], T2 is deduced to (3 | 1 | 2)[])

detailed explanation

In some perspective, unknown could be recognized as type of all types. Therefore, [unknown] is type of all type-level 1-tuples and intuitively, unknown[] is type of all type-level n-tuples.

So [] <: unknown[] is true (i.e. all empty type-level tuple is subtype of all type-level n-tuples) and you can check it simply with this code:

type M = [] & unknown[] extends [] ? [] extends [] & unknown[] ? true : false : false; // true

However, in spite of the fact that [] is subtype of unknown[] therefore [] & unknown[] is actually [], variadic element of [] & unknown[] somehow spreads as never[], which is incorrect and unintuitive.

Playground Link: Playground

Beginne damit, die TypeScript-4.0.3-Beispiele im verlinkten Playground zu reproduzieren, und vergleiche die inferierten Typen für T1 und T2. Die Arbeit ist abgeschlossen, wenn die variadischen Elemente der gezeigten tupelartigen Schnittmengen die erwarteten Tupelformen beibehalten, einschließlich [] und [1, 2, 3].