Give me a concrete example of an overload you want to write but wouldn't be able to if `TypeGaurd` is considered a subtype of bool by type checkers.

(again regardless of what type checkers think typeguard variables will be bools so `isinstance(guard, bool)` is going to be True and `isinstance(guard, TypeGuard)` is going to raise TypeError).

On Wed, Mar 17, 2021 at 5:41 PM Jake Bailey via Typing-sig <typing-sig@python.org> wrote:

Except then no existing function (like filter) could ever get a new overload with TypeGuard, because the compatibility would make it illegal, reducing it's usefulness.
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