ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 27 Jun 2016 18:06:30 +0200Solving Logic Problemshttps://ask.sagemath.org/question/33935/solving-logic-problems/ A few years ago I asked [this](http://ask.sagemath.org/question/10068/solving-logic-problems/) question. I have another question along the same lines. Sorry about the <-> notation, I don't know how else to indicate bijection:
children = { Abe, Dan, Mary, Sue }
ages = { 3, 5, 6, 9 }
children <-> ages #bijection - one child per one age
Abe > Dan #Abe is older than Dan
Sue < Mary #Sue is younger than Mary
Sue = Dan + 3 #Sue's age is Dan's age plus 3 years
Mary > Abe #Mary is older than Abe
Can sagemath determine that:
Abe = 5
Dan = 3
Mary = 9
Sue = 6coder0xffMon, 27 Jun 2016 18:06:30 +0200https://ask.sagemath.org/question/33935/Solving logic problemshttps://ask.sagemath.org/question/10068/solving-logic-problems/Given a set of rules, for example: 1. Mary is older than Tom, 2. Tom is older than Sue; Can sage solve the question, "is Mary older than Sue?"
More specifically, is Sage able to do what Prolog does - unification of logic problems? Thankscoder0xffSat, 27 Apr 2013 15:48:08 +0200https://ask.sagemath.org/question/10068/