Use of Three-State Electronic Level to Express Belief

From: George Levy <glevy.domain.name.hidden>
Date: Tue, 28 Sep 2004 11:44:12 -0700

I am still working to express Lob's formula using the simplest possible
electronic circuit. I am trying to use the well known three-state
concept in electronic as a vehicle for expressing belief .

Let's first define the operator B as a binary operator that uses two
arguments and has one result. Thus the expression qBp means that if q
is 1 then p is known, and and if q is 0 then p is unknown. i.e: qBp ==
If q then p.

Physically this can be implemented by using three-state electronic
technology. According to this technique, an electrical line can be
defined by two voltage levels (eg., 1 and 0) and two impedances (eg.,
HIGH and LOW). Thus an electrical line can have three states:

1) a LOW impedance ON state with a low voltage symbolized by 0
2) a LOW impedance ON state with a high voltage symbolized by 1
3) a HIGH impedance OFF state for "unknown" and symbolized by x.
Physically x could be an arbitrary voltage level other than the ones
assigned for 0 and 1. If a high impedance line is in contact with a low
impedance line the low impedance line dominates.

The truth table for qBp is

q p qBp
0 0 x
0 1 x
1 0 0
1 1 1


AND and OR can easily be defined in terms of 0, 1 and x for two
propositions p and q

AND
p q pq
0 0 0
0 1 0
0 x 0
1 0 0
1 1 1
1 x x
x 0 0
x 1 x
x x x

OR
p q p+q
0 0 0
0 1 1
0 x x
1 0 1
1 1 1
1 x 1
x 0 x
x 1 1
x x x

 For a digital implementation it is necessary to express "implication"
in terms of logical operators using AND, OR , NOT operators.
In general we can convert implication p -> q to a digitally impementable
form: -p + q.
Now let's convert Lob's formula in terms of AND, OR and NOT operators.
Originally Lob's formula is B(Bp -> p) -> Bp.

Since we have defined B as a binary operator we must specify what its
inputs are. Let the left input for the first B be b1 and that for the
second B be b2.
Lob's formula becomes
b1B(b2Bp -> p) -> b1Bp
 

Accordingly, Lob's formula is: ~b1B(~(b2Bp)+ p) + b1Bp

The truth table is

b2 b1 p b1Bp ~(b1Bp)+ p ~b2B(~(b1Bp)+ p)
~b2B(~(b1Bp)+ p) + b1Bp


0 0 0 x x x
                       x
0 0 1 x 1 x
                       x
0 1 0 0 1 x
                       x
0 1 1 1 1 x
                    1
1 0 0 x x x
                     x
1 0 1 x 1 0
                    x
1 1 0 0 1 0
                    0
1 1 1 1 1 0
                    1


I am not sure where this is leading but here it is.

George
Received on Tue Sep 28 2004 - 14:45:36 PDT