On Sat, Feb 04, 2017 at 02:17:44PM -0500, Chris Marshall wrote: > I think $P(a) != $P(b) will work. > > It would probably work with $PDL(a) == $PDL(b) as well but I don't know. > > A quick test should answer the question. I'm hoping Craig or > another PDL::PP expert can weigh in from their experience and > expertise. Thanks, I'll test these solutions. Regards, Luis
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