# Use of Three-State Electronic Level to Express Belief

 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