Equivalent arguments typed with tuples aren't assignable to the same functions with overloads when the same function is recursively referenced

issue にリンクされている TypeScript Playground の再現から始め、タプル引数の関数とオーバーロードされた関数の間で再帰的な代入可能性がどのように動作するかを調査します。再帰的なシグネチャに対する正しいチェックを維持したまま、示されている代入がエラーなしで成功すれば作業完了です。

Bug Report

🔎 Search Terms

arguments, overloads, function, tuple

🕗 Version & Regression Information

All versions with support for tuple arguments (upto v4.5.0-dev.20210815).

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about function overloading (there is nothing in the FAQ about tuples)

⏯ Playground Link

Playground link with relevant code

💻 Code

type START = 0
type DATA = 1

type Args<In, Out> =
  | [t: START, other: FnWithArgs<Out, In>]
  | [t: DATA]

type FnWithArgs<In, Out> = (...args: Args<In, Out>) => void

interface FnWithOverloads<In, Out> {
  (t: START, other: FnWithOverloads<Out, In>): void;
  (t: DATA): void;
}

declare const withArgs: FnWithArgs<1, 2>

// these should be assignable to one another, but aren't:
const assertion: FnWithOverloads<1, 2> = withArgs

🙁 Actual behavior

Two types that should be functionally equivalent aren't assignable to one another:

Type 'FnWithArgs<1, 2>' is not assignable to type 'FnWithOverloads<1, 2>'.
  Types of parameters 'args' and 't' are incompatible.
    Type '[t: 0, other: FnWithOverloads<2, 1>]' is not assignable to type 'Args<1, 2>'.
      Type '[t: 0, other: FnWithOverloads<2, 1>]' is not assignable to type '[t: 0, other: FnWithArgs<2, 1>]'.
        Type at position 1 in source is not compatible with type at position 1 in target.
          Type 'FnWithOverloads<2, 1>' is not assignable to type 'FnWithArgs<2, 1>'.
            Types of parameters 't' and 'args' are incompatible.
              Type 'Args<2, 1>' is not assignable to type '[t: 0, other: FnWithOverloads<1, 2>]'.
                Type '[t: 1]' is not assignable to type '[t: 0, other: FnWithOverloads<1, 2>]'.
                  Source has 1 element(s) but target requires 2.

I've tried simplifying the example, and this issue doesn't seem to occur when the parameters of the function don't reference themselves in their signature.

🙂 Expected behavior

No error, types should be assignable to one another.

Related issues

• https://github.com/microsoft/TypeScript/issues/29732
• https://github.com/microsoft/TypeScript/issues/32164