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Edit detail for #443 equality of nested products is random revision 2 of 4

1 2 3 4
Editor: kratt6
Time: 2009/01/13 16:15:23 GMT-8
Note:

added:
Hm, it seems unreproducible here.   It is reproducible on my machine, though...


changed:
-count(true, [(product(product(f j, j=1..i), i=1..n)=product(product(f k, k=1..j), j=1..n))@Boolean for m in 1..100])
f := operator 'f
count(true, [(product(product(f j, j=1..i), i=1..n)=product(product(f k, k=1..j), j=1..n))@Boolean for m in 1..200])

Submitted by : kratt6 at: 2009-01-13T16:08:03-08:00 (15 years ago)
Name :
Axiom Version :
Category : Severity : Status :
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Hm, it seems unreproducible here. It is reproducible on my machine, though...

axiom
f := operator 'f

\label{eq1}f(1)
Type: BasicOperator?
axiom
count(true, [(product(product(f j, j=1..i), i=1..n)=product(product(f k, k=1..j), j=1..n))@Boolean for m in 1..200])

\label{eq2}200(2)
Type: PositiveInteger?