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Re: fom - 03 - connectivity algebra
Posted:
Dec 7, 2012 5:14 PM
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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
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