Re: Use of Three-State Electronic Level to Express Belief
Bruno I 'believe that the switch analogy is valuable in expressing belief, however, I have trouble making a bridge between this analogy and your explanation. In this post I will make a feeble attempt to make that bridge. To avoid confusion between my Switch belief function and the one you use, let me rename my three state switch belief function from B to S. So now, what I had expressed in an earlier post as qBp becomes qSp, where q is the switch control line, p is the input and qSp is the output. So your Bp becomes my 1Sp and your ~Bp becomes my 0Sp. Note that pSp is never 0. (the control line and the input line carry the same information.) This reminds me of the anthropic principle. Bruno Marchal wrote: Just remember that Bp means the machine has print, or prints, or will print, sooner or later the proposition p. From Bp is true you cannot infer that the machine will print Bp, only that she will print p. Using the switch analogy and paraphrasing what you said: If q = 1 (i.e. Bp is true) then you cannot infer that qSp is true (the machine will print Bp) only that p is known (only that she will pring p) ie., if the control line is on, you still don't know what is the output of the switch. However you know that p is known. Could you pursue this analogy further? George
RE: Use of Three-State Electronic Level to Express Belief
At 08:52 04/10/04 +, Brent Meeker wrote: -Original Message- From: Bruno Marchal [mailto:[EMAIL PROTECTED] Sent: Monday, October 04, 2004 11:52 AM To: Everything List Subject: Re: Use of Three-State Electronic Level to Express Belief You can always try, but let us be sure we agree on the intuitive meaning of Bp, in the case of the Smullyan's self- rererential interpretation of the B. So we have a machine M. The machine M print propositions from time to time. Bp means that the machine M print p. The machine could print Bp. - Could is a slippery word. Is it required that M actually print Bp within a finite number of printings? Just remember that Bp means the machine has print, or prints, or will print, sooner or later the proposition p. From Bp is true you cannot infer that the machine will print Bp, only that she will print p. Even a self-referentially correct machine is not obliged to print Bp when she prints (has printed, will print, ...) p. But a correct machine will never print Bp when she will never print p. Note that a machine can remain consistent and print Bp without ever printing p (this follows from Godel). But here I anticipate perhaps. Now we will say (with Smullyan and the modal logicians) that a machine is normal if the machine prints (has printed, will ...) Bp when she prints (has printed, will ...) p. That machine has some introspective power: when she prints p, she acknowledges the fact: she prints Bp. Obviously such a normal machine will in that case also prints BBp, BBBp, p, Bp, BBp, BBBp, ... You can describe a normal machine by saying that the propositions, Bp-BBp, with p being any propositions, are true about the machine. Again this does not mean the machine will print Bp-BBp for all p. This follows again from the fact that a machine can be normal without printing (communicating, believing, ...) she is normal. OK? Bruno http://iridia.ulb.ac.be/~marchal/
Re: Use of Three-State Electronic Level to Express Belief
At 11:59 29/09/04 -0700, George Levy wrote: Bruno Marchal wrote: Hi George, [out-of-line message] 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 is 0 then p is unknown. How do you manage the known notion.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 qBp. 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. Giving any logic L1, it is always interesting to look if there is no other (perhaps better known) logic L2 such that you can interpret L1 in L2. Now it can be shown that most modal logic cannot be easily or directly represented by a multi-valued logic, so I doubt your proposal could work. You can always try, but let us be sure we agree on the intuitive meaning of Bp, in the case of the Smullyan's self- rererential interpretation of the B. So we have a machine M. The machine M print propositions from time to time. Bp means that the machine M print p. The machine could print Bp. In that case the machine prints the proposition that she prints p. A machine is self-referentially correct (SRC) if her use of B is correct. Examples: The following machine (programs) is SRC: Begin print hello print B hello End The following machine (programs) is not SRC: Begin print B hello End because the machine pretend that she print hello, but will never do it. OK? Of course, we will add conditions; mainly that the machine' set of proposition will be closed for modus ponens, i.e. that if she print (one day, soon or later) X, and if she prints X-Y, then she will print Y. Etc. To sum up: Bp means M asserts p. Bp is true (resp. false) if and only if M asserts p (resp. does not assert p). Bruno http://iridia.ulb.ac.be/~marchal/
Re: Use of Three-State Electronic Level to Express Belief
Bruno Marchal wrote: Hi George, [out-of-line message] 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 is 0 then p is unknown". How do you manage the "known" notion. 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 qBp. 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. George 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 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 http://iridia.ulb.ac.be/~marchal/
