Talk:Product of Summations is Summation Over Cartesian Product of Products

Happy for it to be merged with Product of Summations. It is similar.

Peter Driscoll (talk) 03:46, 10 September 2017 (EDT)


 * I need to analyse it to see whether this approach can genuinely be considered a different proof from the existing one. If they are essentially the same, we may regrettably have to ditch this, as the effort to bring it to house style may be considerable. --prime mover (talk) 04:00, 10 September 2017 (EDT)


 * Note that the statements of the two are not identical. In Product of Summations, $B = \{1\cdots m\}$ does not depend on $a \in A$. So it is a proper generalisation and ought to be treated as such. &mdash; Lord_Farin (talk) 05:12, 10 September 2017 (EDT)