Unknown `DefaultNamedArg` raises type incompatibility errors

🐛 Bug Report

DefaultNamedArg may cause signature compatibility errors. Observed when a function has a default keyword argument in signature and is passed to be used as a function without that particular keyword argument.

To Reproduce

from typing import List, AnyStr
from glob import glob
import itertools as it
def expand_patterns(
        pats: List[AnyStr],
        sort=True
):
    paths = it.chain(map(glob, pats))
    # Argument 1 to "map" has incompatible type "Callable[[AnyStr, DefaultNamedArg(bool, 'recursive')], List[AnyStr]]"; expected "Callable[[AnyStr], List[AnyStr]]"
    if sort:
        paths = sorted(paths)
    return paths

Expected Behavior

Not sure if this ought to happen, but I'd expect a function like foo(x: int, y='y') to be peacefully passable to a function expecting Callable[[int], Any].

Actual Behavior

# Argument 1 to "map" has incompatible type "Callable[[AnyStr, DefaultNamedArg(bool, 'recursive')], List[AnyStr]]"; expected "Callable[[AnyStr], List[AnyStr]]"

Your Environment

• Mypy version used: 0.782
• Mypy command-line flags: Default of VSCode
• Mypy configuration options from mypy.ini (and other config files): N/A
• Python version used: 3.7.7
• Operating system and version: Windows 10, 2004

まず、glob、map、Callable の互換性、DefaultNamedArg を含む、提供された Python の例を再現します。mypy が Callable のシグネチャをどのように比較するかを追跡し、デフォルトのキーワード引数が意図された互換性ルールに従って処理されること、そして報告された非互換性が解消されることを確認します。