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

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