Bruno Marchal wrote:
George, [out-of-line message]
Imagine a three port device such as an electrically controlled switch.
Let's say that this device has three lines connected to it: an input
connected to p, a control connected to q and an output that we'll call
perhaps you could try to motivate your "qBp == If q then p".
I don't see the relation with "if q is 1 then p is known, and and if q
then p is unknown". How do you manage the "known" notion.
If the control sets the switch to OFF (ie. q=0) , the output is not
connected to the input. Therefore for anyone observing the output, the
value of p is unknown, i.e., qBp = x. The electronic value of x can be
any arbitrary value except 0 and 1 which are reserved for the possible
known binary values.
If the control sets the switch to ON (ie. q=1), the output is connected
to the input. Therefore for anyone observing the output, the value of p
is known. It is either 0 or 1 depending on what the input p is.
At 11:44 28/09/04 -0700, you wrote:
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
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
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
0 0 1 x 1 x
0 1 0 0 1 x
0 1 1 1 1 x
1 0 0 x x x
1 0 1 x 1 0
1 1 0 0 1 0
1 1 1 1 1 0
I am not sure where this is leading but here it is.
- Re: Use of Three-State Electronic Level to Express Belief George Levy