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). AND
p(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). OR
p(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 program

pn(pm,ps)

on all 4 inputs and check the result
against the 4 inputs, if there is a duplicate
it can be reduced.

e.g.

and(if(A,B),if(B,C))

the result is the same as

if(A,C)

so I could detect B is eliminated
and a reduction exists.


Herc