Phil Carmody on Tue, 17 Jul 2012 19:16:37 +0200


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Re: (Mod(1,2)*x*y)^0 

--- 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