Allow returning `AnyOf` rather than `Any` for amiguous overloads.

從 overload 規範的第 5 步以及 issue 中連結的 equivalent 詞彙表定義開始。分析 A[Any] 和 A[None] 範例，並判斷有歧義的 overload 是否應保留 union，而不是變成 Any。完成的標準是 typing 規範已針對這種情況制定並記錄明確的規則，且範例反映該規則。

Currently, the typing spec for overloads, step 5, states that

Once this filtering process is applied for all arguments, examine the return types of the remaining overloads. If these return types include type variables, they should be replaced with their solved types. If the resulting return types for all remaining overloads are equivalent, proceed to step 6.

If the return types are not equivalent, overload matching is ambiguous. In this case, assume a return type of Any and stop.

However, couldn't this rule be relaxed to returning the union of the return type of all matching overloads? For instance, consider this example derived from a real world use case (numpy scalar ops)

from typing import Any, overload, assert_type

class A[T]:  # covariant
    def get(self) -> T: ...

@overload
def op(l: A[None], r: A[None]) -> A[None]: ...
@overload
def op(l: A[None], r: A[Any]) -> A[None]: ...
@overload
def op(l: A[Any], r: A[None]) -> A[None]: ...
@overload
def op(l: A[Any], r: A[Any]) -> A[Any]: ...

def test(x: A[None], y: A[Any]) -> None:
    assert_type(op(x, x), A[None])  # spec: ✅️ 
    assert_type(op(x, y), A[None])  # spec: ✅️
    assert_type(op(y, x), A[None])  # spec: ✅️
    assert_type(op(y, y), A[Any] | A[None])     # spec: ❌️ (expected Any)

According to the rule stated above, op(A[Any], A[Any]) should be inferred to as Any, because the overloads are ambious and A[Any] is not equivalent to A[None] as not all materializations of A[Any] can be assigned to A[None].

However inferring Any loses deducible information: we know that no matter what, the return type must be an A. So in principle the return type should be A[Any] | A[None], because if either the left argument or the right argument materializes to A[None], then we get A[None] and otherwise if they materialize to something else we get A[Any].