Literal[False] + Literal[True] does not full cover bool type?

[Holt59]

Describe the bug

When a function has two overloads, one taking Literal[True] and the other Literal[False], pyright throws an error when trying to call it with a bool value or even with a Literal[True, False] value.

Changing bool in the implementation of f1 below to Literal[True, False] does not change the behavior. Using non-bool literals does not produce the issue.

Code or Screenshots

from typing import Literal, overload

@overload
def f1(c: Literal[False] = ...) -> str: ...

@overload
def f1(c: Literal[True] = ...) -> int: ...

def f1(c: bool = False) -> str | int: ...

def r_bool() -> bool: ...
def r_literal() -> Literal[True, False]: ...

f1(r_bool())  # no overload matches for f1
f1(r_literal())  # no overload matches for f1

VS Code extension or command-line

Both, pyright CLI is 1.1.367

I do not remember seeing that problem before so maybe it is a recent update.

[erictraut]

This is a duplicate of #5230.

[Holt59]

This still raises an error and there is no use of bool

from typing import Literal, overload

def f1(c: Literal[False] = ...) -> str: ...
@overload
def f1(c: Literal[True] = ...) -> int: ...
def f1(c: Literal[True, False] = ...) -> str | int: ...

def r_literal() -> Literal[True, False]: ...

f1(r_literal())

In the linked issue you said

Pyright doesn't expand bool into Literal[True, False] in this case. That's an intentional decision, as expansion of this sort would be computationally expensive in a hot path within the type analyzer.

Which I understand, but if you do not expand bool into Literal[True, False], why collapse Literal[True, False] into bool (in the above exxample r_literal() is deduced to return bool)?