[Python-Dev] Rich comparison confusion
Eric S. Raymond
Wed, 17 Jan 2001 19:32:53 -0500
Moshe Zadka <firstname.lastname@example.org>:
> I think that you're confused between two meanings of inverses.
> You think:
> op is an inverse of op' if for every a,b (a op b) = not (a op' b)
> Guido meant (and I hope, implemented):
> op is an inverse of op' if for every a,b (a op b) = (b op' a)
I thought the same.
<pedantic role="defrocked mathematician">
if (a op1 b) <=> (b op2 a), op2 is properly described as the "reflection"
of op1, and vice-versa.
<a href="http://www.tuxedo.org/~esr/">Eric S. Raymond</a>
Sometimes the law defends plunder and participates in it. Sometimes
the law places the whole apparatus of judges, police, prisons and
gendarmes at the service of the plunderers, and treats the victim --
when he defends himself -- as a criminal.
-- Frederic Bastiat, "The Law"