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Phil Carmody on Tue, 17 Jul 2012 19:16:37 +0200
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- To: pari-dev@pari.math.u-bordeaux.fr
- Subject: Re: (Mod(1,2)*x*y)^0
- From: Phil Carmody <thefatphil@yahoo.co.uk>
- Date: Tue, 17 Jul 2012 18:16:25 +0100 (BST)
- Delivery-date: Tue, 17 Jul 2012 19:16:37 +0200
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- In-reply-to: <20120717121522.GC2783@yellowpig>
--- On Tue, 7/17/12, Bill Allombert wrote:
> ? (Mod(1,2)*x*y)^0
> %3 = 1
>
> Is it intended ?
Violating distributivity doesn't sound like a good idea:
? Mod(1,2)^0*x^0*y^0
Mod(1, 2)
Simpler expressions show the loss of type too:
? (Mod(1,2)*x)^0*y
y
Phil