```Date: Dec 7, 2012 5:14 PM
Author: Graham Cooper
Subject: Re: fom - 03 - connectivity algebra

On Dec 7, 8:24 pm, fom <fomJ...@nyms.net> wrote:> >>> >> NOR (NOR,NOR) = OR>> ((p NOR q) NOR (p NOR q)>>    T..F..T...T...T..F..T>    T..F..F...T...T..F..F>    F..T..F...F...F..T..F>    F..F..T...T...F..F..T>>   p  q | p OR q> ------|-------->   T  T |   T>   T  F |   T>   F  F |   F>   F  T |   T>OK, here are the 16 predicates..p(n,A,B,_)p(1,0,0,0).p(1,0,1,0).p(1,1,0,0).p(1,1,1,0).p(2,0,0,0).      ANDp(2,0,1,0).p(2,1,0,0).p(2,1,1,1).p(3,0,0,0).p(3,0,1,0).p(3,1,0,1).p(3,1,1,0).p(4,0,0,0).p(4,0,1,0).p(4,1,0,1).p(4,1,1,1).p(5,0,0,0).p(5,0,1,1).p(5,1,0,0).p(5,1,1,0).p(6,0,0,0).p(6,0,1,1).p(6,1,0,0).p(6,1,1,1).p(7,0,0,0).p(7,0,1,1).p(7,1,0,1).p(7,1,1,0).p(8,0,0,0).    ORp(8,0,1,1).p(8,1,0,1).p(8,1,1,1).p(9,0,0,1).p(9,0,1,0).p(9,1,0,0).p(9,1,1,0).p(10,0,0,1).p(10,0,1,0).p(10,1,0,0).p(10,1,1,1).p(11,0,0,1).p(11,0,1,0).p(11,1,0,1).p(11,1,1,0).p(12,0,0,1).p(12,0,1,0).p(12,1,0,1).p(12,1,1,1).p(13,0,0,1).p(13,0,1,1).p(13,1,0,0).p(13,1,1,0).p(14,0,0,1).p(14,0,1,1).p(14,1,0,0).p(14,1,1,1).p(15,0,0,1).p(15,0,1,1).p(15,1,0,1).p(15,1,1,0).p(16,0,0,1).p(16,0,1,1).p(16,1,0,1).p(16,1,1,1).I might run a programpn(pm,ps)on all 4 inputs and check the resultagainst the 4 inputs, if there is a duplicateit can be reduced.e.g.and(if(A,B),if(B,C))the result is the same asif(A,C)so I could detect B is eliminatedand a reduction exists.Herc
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