带有xor / xnor的Python Sympy bool_map错误？

Question

Python3.5 Sympy似乎认为Xor和Xnor是等效的？

python3.5中的以下代码：

from sympy import *
A1,A2 = symbols('A1,A2')
f1 = Xor(A1,A2)
f2 = ~(Xor(A1,A2))
print(bool_map(f2,f1))
print(bool_map(f1,f2))

输出结果：

((A1 & A2) | (~A1 & ~A2), {A1: A1, A2: A2})
((A1 & ~A2) | (A2 & ~A1), {A1: A1, A2: A2})

所以f1和f2的简化逻辑明显不同，但是bool_map仍然返回它，认为这是2个布尔方程的有效符号映射吗？

我在做什么错了？

编辑：这确实是一个错误

这是_finger 5项指纹的缺陷：

'''
Assign a 5-item fingerprint to each symbol in the equation:
[
# of times it appeared as a Symbol,
# of times it appeared as a Not(symbol),
# of times it appeared as a Symbol in an And or Or,
# of times it appeared as a Not(Symbol) in an And or Or,
sum of the number of arguments with which it appeared,
counting Symbol as 1 and Not(Symbol) as 2
]
'''
from sympy import *
from sympy.logic.boolalg import _finger
from collections import defaultdict
from pprint import pprint

A1,A2 = symbols('A1,A2')
a = _finger((A1 & A2) | (~A1 & ~A2))
b = _finger((A1 & ~A2) | (~A1 & A2))

pprint(a)
pprint(b)

对A1和A2使用相同的指纹：

defaultdict(<class 'list'>, {(0, 0, 1, 1, 6): [A1, A2]})
defaultdict(<class 'list'>, {(0, 0, 1, 1, 6): [A1, A2]})

这就是为什么A1和A2看起来可互换的原因