Regression: Type of instance variable narrowed incorrectly after `is not` expression

Bug Report

Type of a.foo in the example below is narrowed to what looks like a function type which results in a strange error on the call to append later.

Issue started with https://github.com/python/mypy/commit/ad30be89f146adaae398bd854146952158a53cad @hauntsaninja

To Reproduce

from typing import Any

def f(a: Any, x: Any) -> None:
    if x is not list:
        x = []
    reveal_type(x) # note: Revealed type is "Any"
    if a.foo is not list:
        a.foo = []
    reveal_type(a.foo) # note: Revealed type is "Any | Overload(def [_T] () -> builtins.list[_T`1], def [_T] (typing.Iterable[_T`1]) -> builtins.list[_T`1])"
    a.foo.append(x) # error: Too few arguments for "append" of "list"  [call-arg]

Expected Behavior

a.foo should probably be typed as Any.

Actual Behavior

/Users/psawicki/test.py:6: note: Revealed type is "Any"
/Users/psawicki/test.py:9: note: Revealed type is "Any | Overload(def [_T] () -> builtins.list[_T`1], def [_T] (typing.Iterable[_T`1]) -> builtins.list[_T`1])"
/Users/psawicki/test.py:10: error: Too few arguments for "append" of "list"  [call-arg]

Your Environment

• Mypy version used: mypy 1.20.0+dev.ad30be89f146adaae398bd854146952158a53cad and up
• Mypy command-line flags: mypy ~/test.py
• Mypy configuration options from mypy.ini (and other config files): None
• Python version used: 3.13.5

Führe zunächst den bereitgestellten Reproducer mit mypy aus und vergleiche die angezeigten Typen für x und a.foo. Verfolge den Type-Narrowing-Pfad für den Ausdruck a.foo is not list. Als erledigt gilt, wenn a.foo weiterhin als Any typisiert bleibt und der spätere Aufruf append(x) den gezeigten Argumentfehler nicht mehr meldet.