missing 'xor' Boolean operator

[Terry Reedy]

Albert van der Horst wrote:

> Remains whether we need an xor that only works and requires that
> both operands are booleans. That one we have already!
> It is called != .
> 
> (a!=b)!=c
> and
> a!=(b!=c)
> 
> are the same for booleans, so can indeed be expressed
> a!=b!=c   (associativy of xor)

Not in Python

 >>> a,b,c = True, False, True
 >>> (a!=b)!=c
False
 >>> a!=(b!=c)
False
 >>> a!=b!=c
True
 >>> (a!=b) and (b!=c)
True

In Math and Python, a<b<c means a<b and b<c, not (a<b)<c or a<(b<c).
!= is a comparison operator like <, not an arithmetic operator like + 
(or logical xor). If one has 0/1 or True/False values, we have 
arithmetic xor already, ^, which works as expected.

 >>> (a^b)^c
False
 >>> a^b^c
False

Terry Jan Reedy