Re: ... cosmology? KNIGHT KNAVE
At 09:53 29/07/04 -0700, Hal Finney wrote: Tell me again where I am going wrong. OK. Consider each of these examples: 117. q ... 191. Bp ... 207. p - q Now, we will say that the machines believes something if it is one of its theorems, right? So we can say that the machine believes q, it believes Bp, and it believes p-q, right? We could equivalently say it believes q is true, etc., but that is redundant. If it writes x down as a theorem, we will say it believes x, which is shorthand for saying that it believes x is true. The is true part has no real meaning and does not seem helpful. We also have this shorthand Bx to mean the machine believes x. So we (not the machine, but us, you and I!) can also write, Bq, BBp and B(p-q), and all of these are true statements, right? Until here you are right. The problem arises when we start to use this same letter B in the machine's theorems. It is easy to slide back and forth between the machine's B and our B. But there is no a priori reason to assume that they are the same. That is something that has to be justified. The problem arises here, indeed. And I disagree with what you are saying. We take, by personal choice, the total NAIVE STANCE toward the machine. That means that by definition Bx means for us, and for the machine that the machine believes x. Exactly like when the machine believes (p - q), it means, like for us (-p v q), that p is false or q is true. If the machine believes Bp, it just means that the machine believes it will believe p. In case the machine print Bp and then never print p, it will means (for us) that the machine has a false belief. Focus on 207. p-q for a moment. We know that, according to the machine's rules, this theorem means that if it ever writes down p as a theorem, it will write down q. Take care. p-q is also true in case p is false (by propositional logic), it could means that the machine believes -p. Let us look at the following example: with f denoting any contradiction (that is f can be seen as an abbreviation of (p -p)) The machine obeys to classical propositional calculus (CPC). Thus the machine believes all proposition f-p (with any p). And what you said is correct, that will entails that if the machine believes f, then it will believe p, and then, if we add the assumption that the machine is consistent, it will never believes f, that is Bf is false for the machine, and so we know then that [Bf - what-you want] will always be true (because we know also the CPC. Therefore it is true that Bp-Bq. Yes. Because the machine obeys CPC. This is simply another way of saying the same thing! Bp means that p is a theorem, by definition of the letter B, in the real world. And similarly Bq means that q is a theorem. Given that p-q is a theorem, then if p is a theorem, so is q. Therefore it is true that B(p-q) - (Bp - Bq). This is not a theorem of the machine, it is a truth in the real world. Right. What I want to say is that 207. p-q means Bp - Bq. It means that if the machine ever derives p, it will derive q. This is a true statement about the operations of the machine. It is not a theorem of the machine. OK, but what is a theorem by the machine, is p-q, independently that *we* know this entails Bp - Bq. So your expression p-q means Bp - Bq, is misleading. The naive stance is that the machine believes p-q. (or, if we want to insist, that the machine believes p-q is true, but as you said this does not add anything. Actually p-q could be false, and the machine could have false beliefs, in which case both p-q and (p-q) is true are false. When we talk about what something means, I think it has to be what it means to us, not what it means to the machine. Why? If you talk with any platonist, you should better keep the naive stance. If not it is like you suspect some problem with the platonist brain, and you will no more talk about the same thing with him. It will be much harder for you to show him that he is doing a mistake for exemple. When the machine writes 117.q, it doesn't mean anything to the machine. Why? With the naive stance, it means the machine believes q. For perhaps unknown reason to us the machine believes q. Perhaps the machine has make a visit to the KK island; and some native told her that if I am a knight then q. Or more simply the native said 1+1 = 2, and then latter said q. I have said that the machine believes all classical tautologies, and that if the machine believes X, and then X-Y, then the machine believes Y. But I NEVERsaid that the machine believes ONLY the tautologies. The machine can have its personal life and got some personal non logical beliefs (non tautological belief) on its own. Like in some of the problem I gave where machine develops beliefs on the knight/kanve nature of the natives. To us it means that the machine believes q, or that the machine believes q is true. This is right, but keep in mind that it means something for the machine. It means q, or q is
Re: ... cosmology? KNIGHT KNAVE
At 12:47 30/07/04 +0200, I wrote: Oh, any accurate machine (for which Bp-p is true) is obviously normal. This is false. But an accurate stable machine will be stable. Just substitute p with Bp in (Bp - p) to get BBp - Bp. That's stability, not normality. Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
Hi John, At 17:19 26/07/04 -0400, John M wrote: Bruno, (and ClassG) We have an overwhelming ignorance about Ks and Ks. We don't know their logical built, their knowledege-base, their behavior. Indeed. Is the K vs K rule a physical, or rather human statement, when - in the latter case there may be violations (punishable by jail - ha ha). Neither physical, nor human ... (see below). Do K K abide by 100.00% by the ONE rule we know about them, or ~99.999%, when there still may be an aberration? 100,00% Are they robots or humans? Looks like machines. Are machines omniscient? Interesting question (not addressed by Smullyan!). But easy though. From Godel's incompleteness (which we have not yet proved, except in the diagonalisation post some time ago, but on which we will come back: it is the heart of the matter in FU's term), it will be easy to prove that: - Machine cannot be omniscient. - Both knight and knaves are omniscient, and so they cannot be machine. I expect, but will not argue now, that knights cannot exist at all, even in platonia (and this with or without comp). Does this throws doubts on what we can infer from FU's puzzles? No, because the KK island is just a pedagogical tool for building a fictive but easily imaginable situation where reasoners must believe some self-referential propositions. But with the diagonalization lemma (alias the heart of the matter) we will eliminate the need of the KK island. It is the logical fate of the correct machine to meet inescapably true and believable (provable) self-referential propositions, from which we can derive true but unbelievable propositions, ... and much more. Bruno PS: Thanks to those who have send me hard puzzles! I will try to solve them after 16 August. I will be busy until then. I will just answer Hal Finney KK Posts, and then finish my paper. I hope I will get the authorization to make it public soon for it will be a good base to proceed on. It is a step toward the English paper I promised to Wei Dai, a long time ago. http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
At 09:54 27/07/04 -0700, Hal Finney wrote: I am confused about how belief works in this logical reasoner of type 1. Suppose I am such a reasoner. I can be thought of as a theorem-proving machine who uses logic to draw conclusions from premises. We can imagine there is a numbered list of everything I believe and have concluded. It starts with my premises and then I add to it with my conclusions. OK. In this case my premises might be: 1. Knights always tell the truth 2. Knaves always lie 3. Every native is either a knight or a knave 4. A native said, you will never believe I am a knight. Now we can start drawing conclusions. Let t be the proposition that the native is a knight (and hence tells the truth). Then 3 implies: 5. t or ~t Point 4 leads to two conclusions: 6. t implies ~Bt 7. ~t implies Bt Here I use ~ for not, and Bx for I believe x. I am ignoring some complexities involving the future tense of the word will but I think that is OK. Perfect. Here Hal believes p means that sooner or later Hal will assert, believe or prove p. It means p belongs to the list you mentionned. However now I am confused. How do I work with this letter B? What kind of rules does it follow? I understand that Bx, I believe x, is merely a shorthand for saying that x is on my list of premises/conclusions. Correct. This means Bx is a equivalent with Hal believes x. The only difference is that Bx is supposed to be in the language of the machine. If I ever write down x on my numbered list, I could also write down Bx and BBx and BBBx as far as I feel like going. Is this correct? Well, not necessarily. Unless you are a normal machine, which I hope you are! So let us accept the following definition: a machine is normal when, if it ever assert x, it will sooner or later asserts Bx. Normality is a form of self-awareness: when the machine believes x, it will believe Bx, that is it will believe that it will believe x. But what about the other direction? From Bx, can I deduce x? That's pretty important for this puzzle. If Bx merely is a shorthand for saying that x is on my list, then it seems fair to say that if I ever write down Bx I can also write down x. But this seems too powerful. You are right. It is powerful, but rather fair also. let us define a machine to be stable if that is the case. When the machine believes Bx the machine believes x. So what are the correct rules that I, as a simple machine, can follow for dealing with the letter B? Actually we will be interested in a lot of sort of machine. But I do hope you are both normal and stable. Actually I'm sure you are. The problem is that the rules I proposed here lead to a contradiction. If x implies Bx, then I can write down: 8. t implies Bt Note, this does not mean that if he is a knight I believe it, but rather that if I ever deduce he is a knight, I believe it, which is simply the definition of believe in this context. Here you are mistaken. It is funny because you clearly see the mistake, given that you say 'attention (t implies Bt) does not mean if he is a knight the I believe it. But of course (t implies Bt) *does* mean if he is a knight the I believe it. You add it means only (and here I add a slight correction) if I ever deduce he is a knight I will deduce I believe he is a knight which really is, in the machine language: (Bt implies BBt) instead of (t implies Bt). To be sure: a machine is normal if for any proposition p, if the machine believes p, it will believe Bp. But this is equivalent with saying that for any proposition p, the proposition (Bp implies BBp) is true *about* the machine. Same remark for stability: you can say a machine is stable if all the propositions (BBp implies Bp) are true about the machine. This does not mean the stable or normal machine will ever believe being stable or normal. You have (momentarily) confuse a proposition being true on a machine, and being believe by a machine. But 6 and 8 together mean that t implies a contradiction, hence I can conclude: 9. ~t He is a knave. 7 then implies 10. Bt I believe he is a knight. And if Bx implies x, then: 11. t and I have reached a contradiction with 9. So I don't think I am doing this right. By taking into account the confusion above, you should be able to prove, with t still the same proposition (that is (t - -Bt)), that (in case you are a normal reasoner of type 1): if you are consistent, then t is not provable (believable, assertable by you) if you are consistent and stable, then -t is not provable either. That's Godel's first incompleteness theorem. (once we eliminate the KK island from the reasoning to be sure). Bravo. To sum up; any normal stable reasoner of type 1 meeting a knight saying you will never believe I'm a knight will be forever incomplete. (incomplete = there is a proposition like t, which is neither provable nor refutable). And so a machine cannot be omniscient because, although the KK island does not exist, the diagonalization lemma will
Re: ... cosmology? KNIGHT KNAVE
This is confusing because I believe p has two different meanings. One is that I have written down p with a number in front of it, as one of my theorems. The other meaning is the string Bp. But that string only has meaning from the perspective of an outside observer. To me, as the machine, it is just a pair of letters. B doesn't have to mean believe. It could mean Belachen, which is German for believe. All I need to know, as a formal system, is what rules the letter B follows. Bruno wrote: At 09:54 27/07/04 -0700, Hal Finney wrote: If I ever write down x on my numbered list, I could also write down Bx and BBx and BBBx as far as I feel like going. Is this correct? Well, not necessarily. Unless you are a normal machine, which I hope you are! So let us accept the following definition: a machine is normal when, if it ever assert x, it will sooner or later asserts Bx. Normality is a form of self-awareness: when the machine believes x, it will believe Bx, that is it will believe that it will believe x. But what about the other direction? From Bx, can I deduce x? That's pretty important for this puzzle. If Bx merely is a shorthand for saying that x is on my list, then it seems fair to say that if I ever write down Bx I can also write down x. But this seems too powerful. You are right. It is powerful, but rather fair also. let us define a machine to be stable if that is the case. When the machine believes Bx the machine believes x. So in my terms, I can add two axioms: 0a. x implies Bx 0b. Bx implies x The first is the axiom of normality, and the second is the axiom of stability. I don't find these words to be particularly appropriate, by the way, but I suppose they are traditional. It also seems to me that these axioms, which define the behavior of the letter B, don't particularly well represent the concept of belief. The problem is that beliefs can be uncertain and don't follow the law of the excluded middle. If p is that there is life on Mars, then (p or ~p) is true. Either there's life there or there isn't. But it's not true that (Bp or B~p). It's not the case that either I believe there is life on Mars or I believe there is no life on Mars. The truth is, I don't believe either way. But axioms 0a and 0b let me conclude (Bp or B~p). Obviously they collectively imply p if and only if Bp. Therefore from (p or ~p) we can immediately get (Bp or B~p). Hence for normal people, the law of the excluded middle applies to beliefs. This proof is pure logic and has no dependence on the meaning of B. If B is Belachen, I have showed that if p implies Belachen(p), then it follows that (Belachen(p) or Belachen(~p)) is true. That's all. It's a step outside the system to say that B follows rules which make it appropriate for us to treat it as meaning believes. But do 0a. and 0b. really capture the meaning of belief? I question that. B looks more like an identity operator under those axioms. The problem is that the rules I proposed here lead to a contradiction. If x implies Bx, then I can write down: 8. t implies Bt Note, this does not mean that if he is a knight I believe it, but rather that if I ever deduce he is a knight, I believe it, which is simply the definition of believe in this context. Here you are mistaken. It is funny because you clearly see the mistake, given that you say 'attention (t implies Bt) does not mean if he is a knight the I believe it. But of course (t implies Bt) *does* mean if he is a knight the I believe it. I don't see this. To me as the machine, there is no meaning. I am just playing with letters. t implies Bt is only a shorthand for if he is a knight then B(if he is a knight). There is no more meaning than that. The letter B is just a letter that follows certain rules. We only get meaning from outside, when we look at what the machine is doing and try to relate the way the rules work to concepts in the real world. It is at this point that we bring in the interpretation of Bx as the machine believes x. Suppose for some proposition q the machine deduces it on step 117: 117. q Does this mean that q is true? No, it means that that the machine believes q. Does it mean that Bq is true? Yes. Bq is true, because Bq is a shorthand for saying that the machine believes q, and by definition the machine believes something when it writes it down in its numbered list. We can see it right there, number 117. So the machine believes q and Bq is true. But q is not (necessarily) true. The machine writing something down does not mean it is true. By definition, it means the machine believes it. Consider a different example: 191. Bp What does this mean? Does it mean that Bp is true? No, it means that the machine believes Bp, because by definition, what the machine writes down in its numbered list is what it believes. Is BBp true? Yes, it is true, because that says that the machine believes Bp, and that means that Bp is in the
Re: ... cosmology? KNIGHT KNAVE
I am confused about how belief works in this logical reasoner of type 1. Suppose I am such a reasoner. I can be thought of as a theorem-proving machine who uses logic to draw conclusions from premises. We can imagine there is a numbered list of everything I believe and have concluded. It starts with my premises and then I add to it with my conclusions. In this case my premises might be: 1. Knights always tell the truth 2. Knaves always lie 3. Every native is either a knight or a knave 4. A native said, you will never believe I am a knight. Now we can start drawing conclusions. Let t be the proposition that the native is a knight (and hence tells the truth). Then 3 implies: 5. t or ~t Point 4 leads to two conclusions: 6. t implies ~Bt 7. ~t implies Bt Here I use ~ for not, and Bx for I believe x. I am ignoring some complexities involving the future tense of the word will but I think that is OK. However now I am confused. How do I work with this letter B? What kind of rules does it follow? I understand that Bx, I believe x, is merely a shorthand for saying that x is on my list of premises/conclusions. If I ever write down x on my numbered list, I could also write down Bx and BBx and BBBx as far as I feel like going. Is this correct? But what about the other direction? From Bx, can I deduce x? That's pretty important for this puzzle. If Bx merely is a shorthand for saying that x is on my list, then it seems fair to say that if I ever write down Bx I can also write down x. But this seems too powerful. So what are the correct rules that I, as a simple machine, can follow for dealing with the letter B? The problem is that the rules I proposed here lead to a contradiction. If x implies Bx, then I can write down: 8. t implies Bt Note, this does not mean that if he is a knight I believe it, but rather that if I ever deduce he is a knight, I believe it, which is simply the definition of believe in this context. But 6 and 8 together mean that t implies a contradiction, hence I can conclude: 9. ~t He is a knave. 7 then implies 10. Bt I believe he is a knight. And if Bx implies x, then: 11. t and I have reached a contradiction with 9. So I don't think I am doing this right. Hal Finney
Re: ... cosmology? KNIGHT KNAVE
(problem 4) You get a native, and asks her if Santa Claus exists. The native answers this: "If I am a knight then Santa Claus exists" What can you deduce about the native, and about Santa Claus?Lets give a name to the sentence:S="If I am a knight then Santa Claus exists"1. If the native is a knight, then S is true. If S is true and the native isa knight, then Santa Claus exists. Therefore, Santa Claus exists.2. If the native is a knave, then S is false. If S is false (1-0), then thenative must be a knight. So he can't be a knave. So a knave could never say S and be consistent.Conclusion: The native is a knight and Santa Claus exists.Jan
Re: ... cosmology? KNIGHT KNAVE
At 16:15 23/07/04 +0200, Jan Harms wrote: (problem 4) You get a native, and asks her if Santa Claus exists. The native answers this: If I am a knight then Santa Claus exists What can you deduce about the native, and about Santa Claus? Lets give a name to the sentence: S=If I am a knight then Santa Claus exists 1. If the native is a knight, then S is true. If S is true and the native is a knight, then Santa Claus exists. Therefore, Santa Claus exists. 2. If the native is a knave, then S is false. If S is false (1-0), then the native must be a knight. So he can't be a knave. So a knave could never say S and be consistent. Conclusion: The native is a knight and Santa Claus exists. After reading ten times I conclude you got the solution. I have been mislead by the fact that there are mainly two ways to solve the puzzle (and George did try the two ways), and your 1. looks like the beginning of the first way, and your 2. looks like the second way. Also the conclusion of your 1. is that S is true. You still need your 2. to conclude, of course (as you illustrate). And then I have been mislead because it looks you assume a knave to be consistent. But if you ask a knave if 1=0 he answers yes, so he is certainly not consistent in the usual sense of the word (which we will define when we will introduce the believe ...). But I eventually realize you were just meaning by consistent: consistent with his status of knave. So ok you were right. Let me explain again. I give two independent solutions: A and B. I recall, you are on the KK island, and a native tells you If I am a knight then Santa Claus exists. A. Let us suppose the native is a knight. Then what he tells us is true. So if we suppose the native is a knight the *two* following proposition are true: -the native is a knight (because if the native is a knight the native is knight, isn'it?) -S is true, that is: if the native is a knight then Santa Claus exists. from which we can conclude that Santa Claus exists. So we have shown that indeed IF the native is a knight then Santa Claus exists. (Indeed By supposing that the native is a knight we get the existence of Santa Klaus). At this point it is important to see that we have prove only that S is true (not that Santa Claus exists). OK. But the native asserted S, and only a knight can assert true proposition. So the native is a knight. So know we *know* the two propositions are true: -the native is a knight -S is true, that is: if the native is a knight then Santa Claus exists. From which we can conclude that Santa Claus exists. From which we can conclude, BTW, that either a knight knave island does not exist or, if it exists, no native will ever said to you if I am a knight Santa Claus exists, or (in case the KK island exists and a native tells you the S sentence) ... that Santa Claus exists. B. Let us suppose the native is knave. Then what he said was false. But he said if I am a knight then Santa Claus exists. That proposition can only be false in the case he is a knight and Santa Claus does not exists. So if he is a knave, he must be a knight and that's a contradiction. So he cannot be a knave, and so he is a knight. But then what he said was true, and giving what he said, Santa Claus exists. OK? That should give you a smell of Lob. At some point it will be necessary to understand that in computerland the option of saying the KK island does not exist just does not work for hunting away the self-referential proposition. Comp entails some Lobian magic around! But I anticipate. I see you want the next problem. OK. I hope you all agree that a native, not a tourist, a real native from the KK island will never tell you I am a knave, or, it is equivalent I am not a knight. A knight saying that would be lying, a knave saying that would be telling the truth. Now with the next two problems we are doing a big step and a very big step. So big that we will be forced to work again on 1-4 a little bit more formally. Big Problem 5: Could a native tell you You will never know that I am knight ? Very Big Problem 6: Could a native tell you You will never believe that I am knight ? Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
Dear Bruno and Friends, After having read Smullyan's wonderful little book and reading these posts I would like to point out a problem that I see. The notion of Knights and Knaves, as Truth and Falsehood-tellers (or reporters) respectively, tacitly assumes that these entities are Omniscient, e.q., that they have access to a list of all Possible Truths/Falsehoods or what ever is equivalent. No effort seems to be made to explain exactly how it is that this assumption can be related to the actual world of experience, a world where information is finite, oracles are often wrong, perpetual motion is impossible and distributions are almost never Gaussian nor linear. This, I believe, is related to the main problem that I have with the Platonic approach to Logic, Mathematics and COMP (among others), it grants God-like powers to entities - infinite computational resources, infinite heat sinks/sources, etc. - and in so doing allows for us to fool ourselves that very difficult problems, such as found in cosmology, can easily be solved. Kindest regards, Stephen
Re: ... cosmology? KNIGHT KNAVE
Bruno Marchal wrote: Let us suppose the native is knave. Then what he said was false. But he said if I am a knight then Santa Claus exists. That proposition can only be false in the case he is a knight and Santa Claus does not exists. This only works if you assume his if-then statement was shorthand for the logical conditional, -, in formal logic (see http://en.wikipedia.org/wiki/Logical_conditional )...if you interpret it some other way, like that it was shorthand for a modal logic idea like in every possible world where it is true that I am a knight, it is true that Santa Claus exists, I don't think it can only be false if he is a knight. For example, there might be a possible world where he is a knight and Santa Claus does *not* exist, in which case the statement in every possible world where it is true that I am a knight, it is true that Santa Claus exists is false. I think this is why the problem is confusing--for me, possible-world statements more accurately capture the meaning of if-then statements in ordinary language than the logical conditional. Jesse
Re: ... cosmology? KNIGHT KNAVE
Hi George, At 22:17 22/07/04 -0700, George Levy wrote: Hi Bruno Bruno Marchal wrote: You get a native, and asks her if Santa Claus exists. The native answers this: If I am a knight then Santa Claus exists What can you deduce about the native, and about Santa Claus? First let's assume that the native is a knight. Since he tells the truth, then Santa Claus must exist. That's all,... we cannot go any further. Do you see now that we can go further? You just showed true that if he is a knight Santa Claus exists, but that is what he said so he said something true, meaning he *is* a knight and then ... Now let's assume that the native is a knave. Then the statement he made is false. The corresponding true statement is: If I am a knight then Santa Claus does not exist. False statement you mean? I mean p - q is false when p is true and q is false. However we assumed that the native is not a knight. Therefore the statement does not apply. No information can be obtained from this statement. All right somehow you make a point, but, as Stephen deplores, we are in Platonia. Do you agree that, (with x number): for all x, if x is bigger than 10 then x is bigger than 5. If you agree you are in platonia giving that you have accepted that the (admittedly vacuous) truth of all the following propositions: if 1 is bigger than 10 then 1 is bigger than 5 if 6 is bigger than 10 then 6 is bigger than 5 if 100 is bigger than 10 then 100 is bigger than 5 So you accept the truth table of p - q 1 1 1 1 0 0 0 1 1 0 0 0 p - q is the same as -p v q, or -(p -q) So if a *knave* say (A - B), it means really means -(A - B) = (A and -B) (the second row). OK? Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
Bruno, If I am a Knight... is definitely Knave: a Knight would not make it a condition of his stating whether S.Cl is 'true' upon HIS status in the world. Such wishy-washy statement is Knavish. It would be better put so: Since I am... - but still not Knightish (although we did not hear about the logical prowess of Knights). What difference would it make on S.Cl. if 'he' is anything else? Bad proposition I think (PC). John Mikes - Original Message - From: Bruno Marchal [EMAIL PROTECTED] To: Everything List [EMAIL PROTECTED] Sent: Friday, July 23, 2004 9:09 AM Subject: Re: ... cosmology? KNIGHT KNAVE Hi George, At 22:17 22/07/04 -0700, George Levy wrote: (problem 4) You get a native, and asks her if Santa Claus exists. The native answers this: If I am a knight then Santa Claus exists What can you deduce about the native, and about Santa Claus? First let's assume that the native is a knight. Since he tells the truth, then Santa Claus must exist. That's all,... we cannot go any further. Now let's assume that the native is a knave. Then the statement he made is false. The corresponding true statement is: If I am a knight then Santa Claus does not exist. However we assumed that the native is not a knight. Therefore the statement does not apply. No information can be obtained from this statement. We still don't know if the native is a knight or a knave, and we still do not know if Santa exists or not. Does everybody agree with George? Well, if everybody agree then ... everybody should think twice! (It *is* a little bit tricky, but that trickiness is really what we need to understand Godel, Lob, and then, as you can suspect, the derivation of physics from logic/arithmetic through Godel, Lob, ...). For the fun, I let everybody think twice! (Actually you really need to think twice. This is a hint. Another hint: the reasoning toward the solution will look a little bit circular, but only look; appearance can be deceiving. Things will be more tricky when we will introduce the modal know or believe, but that's premature. We are still in pure PC. (PC = Propositional Calculus). George, thanks for your attempt. There is no shame to be wrong of course, on the contrary it is the *only* way to learn. Note that some professional mathematician have criticize my thesis by doing similar error!!! (most acknowledge at time, but not all!!!). It is my revelation of the last ten years, logic is not well known even by scientist. ... and the problem 4 is somehow tricky, I let you enjoy thinking twice. If nobody solves the problem 4, I will give the solution tomorrow, unless someone asks for having more time ... Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
Bruno, Nice story and game depiction; it does help - somewhat - to explain a more expansive generalization of 'decidability' ..the bedrock on which 'logic' (at least for the traditional understanding of that term) relies. Global consistency 'permits' decidability 'which permits' logic. But there are prefaces to -those- relations. And direct indication thereby that consistency is 'necessary' but not alone 'sufficient' to arise 'decision', and then 'logic'. Case: have your student place the call. but the native answers in cantonese ; or, doesn't know the significance of the device called 'phone' and thinks it just an interesting noise-maker. A 'consistency' of co-presence would exist in such a universe, but not a 'requisite interaction' rule. Yet a 'logic of co-existence' -would- exist _strong enough and pervasive enough_ to accomodate co-presence -with- 'involvement and no involvement' simultaneously. So .. first there must be a: global consistency 'which permits' decidability AND non-decidability before you can generate the option sub-path .. Global consistency 'permits' decidability 'which permits' logic. Jamie Rose Ceptual Institute
Re: ... cosmology? KNIGHT KNAVE
James, You may be saying something, but the problems are not that sophisticate. There where default hypothesis, sure, like the hypothesis that the Knights and Knaves understand English ..., knows how to use a phone, and are able to survive more than a nanosecond ... There might be sense in your remarks about consistency, but the puzzles are supposed to introduce slowly the not so easy notion of consistency. Also logic does not necessitate decidability. Let go slowly. Bruno At 06:26 22/07/04 -0700, James N Rose wrote: Bruno, Nice story and game depiction; it does help - somewhat - to explain a more expansive generalization of 'decidability' ..the bedrock on which 'logic' (at least for the traditional understanding of that term) relies. Global consistency 'permits' decidability 'which permits' logic. But there are prefaces to -those- relations. And direct indication thereby that consistency is 'necessary' but not alone 'sufficient' to arise 'decision', and then 'logic'. Case: have your student place the call. but the native answers in cantonese ; or, doesn't know the significance of the device called 'phone' and thinks it just an interesting noise-maker. A 'consistency' of co-presence would exist in such a universe, but not a 'requisite interaction' rule. Yet a 'logic of co-existence' -would- exist _strong enough and pervasive enough_ to accomodate co-presence -with- 'involvement and no involvement' simultaneously. So .. first there must be a: global consistency 'which permits' decidability AND non-decidability before you can generate the option sub-path .. Global consistency 'permits' decidability 'which permits' logic. Jamie Rose Ceptual Institute http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
George,
At 21:17 20/07/04 -0700, George Levy wrote:
Bruno, John, Russell
I am half-way through Smullyan's book.
Nice! You will see how easy it will be to state precisely the main result
and the open problems in my thesis once you grasp the whole FU.
Of course, to really appreciate, there will be a need to study a little bit
two non-classical logics: intuitionist logic and quantum logic ('course).
It is an entertaining book for someone motivated enough to do all these
puzzles, but I think that what is missing is a metalevel discussion of
what all this means.
Sure! I see that now that we are going back to the basic you are
impatiently anticipating the things! That's makes me happy!
Mathematical fireworks occur because we are dealing with self-referential
systems. In the old days they may have called them
reflexive. Reflection is, I think, an essential component of conscious
thought.
Yes.
The type of reflection I have encountered so far in the book involves
infinite reflections which lead to paradoxes.
For example if someone says I am a knave then obviously we have a paradox.
I disagree!!!
I'm a little bit sorry because I have not resisted
putting a little trap in the problems yesterday.
Actually there is no paradox here!
(More below).
The human mind, however, does not have the capacity to deal with an
infinite number of reflections. (I think that you think that I think that
you think.).
You anticipate a little bit too much. In a sense
not only humans can deal with infinite
reflexion but machine can do aswell...I will not try to explain
now. In FU this is explained in the heart of
the matter section 25. To be honest Smullyan's
explanation is rather short. When we will arrive
at that stage it will be good to (re)read the diagonalisation
posts (accessible from my url).
There are two really fundamental theorems: their official
name are
1) The second recursion theorem of Kleene (2-REC)
2) The diagonalisation lemma (DL)
The first one can aptly be called (with the comp hyp) the
fundamental theorem of theoretical biology. It is the one I have use
to build amoeba (self-reproducing programs), planaria (self
regenerating programs) and dreaming machine (programs
capable of conceiving themselves in alternate anticipations).
Solovay's arithmetical completeness theorems of G and G* are
themselves the most pretty application of 2-REC. George Boolos
use the DL instead. They are quite related.
If, however, self referential systems are limited to a finite number of
reflections, such as the human mind is capable of, then these paradoxes
may go away. With one reflection a knave would says: I am a knight.
With two reflections he would say I am a knave. With three reflections,
I am a knight. With four I am a knave. and so on. With an infinite
number of reflections he would remain Forever Undecided.
I am not sure if Physics is derived from an ideal infinite
self-referential systems or from a more human and messy finite system and
I cannot think of an obvious and clear-cut justification for either
approach. What do you think?
Physics will come from the *machine* messy
finite system, not necessarily *human*.
Actually the machine's psychology G and G*
is also the psychology of some infinite machine
and there exists for each sort of alpha-machine
(alpha ordinal) a related physics. So by testing those
physics we will be able to evaluate our own degree
of comp/non-comp. My thesis shows that quantum sort of logic
appears right at the starting point alpha = omega point.
And now the solution of the problems:
(I see most people have find the solutions (and even new problem and
solution, thanks to John and Russell) ... modulo some details.
The native of that Island are all either knight or knaves
and knight always tell the truth, and knaves always lie.
You go there.
Problem 1. A native tell you I am a knight. Is it possible to deduce
the native's type?
Problem 2. You meet someone on the island, and he tells you
I am a knave. What can you deduce?
Problem 1. The answer is NO. The assertion I am a knight
can be made by a knight (who then tells the truth as it should),
but can be made by a knave (who then tells a lie as it should).
as it should giving the rule of the island. OK?
Problem 2. Someone says I am a knave. Suppose it is a
knave: well then he says the truth so it cannot be a knave.
Suppose it a knight: well then it is a lie so it cannot be a knight, either.
So it is neither a knight, nor a knave, but we have been said
that all the native of the island are either knight or knave.
So we can deduce that it cannot be a native of the
island. That someone (the trap!) is most probably
a (lying) tourist, or a mad explorator ... OK?
;-)
Bruno
http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
- Original Message - From: Russell Standish [EMAIL PROTECTED] To: John M [EMAIL PROTECTED] Cc: [EMAIL PROTECTED]; Bruno Marchal [EMAIL PROTECTED] Sent: Tuesday, July 20, 2004 7:29 PM Subject: Re: ... cosmology? KNIGHT KNAVE Russell, your solution (in your attachment) is the right one: the 'double reverse': eliminting the negative by double negation, while the 'double positive' still stays positive. ^^ What would your other brother say is the road to Baghdad? Then take the other direction! Cheers ^ How would that be altered in George Levy's nth 'reflection'? (just kidding). John M
... cosmology? KNIGHT KNAVE
At 09:55 20/07/04 -0400, John Mikes wrote: It all depends what do we deem: POSSIBLE. According to what conditions, belief, circumstances? If we accept the here and now as the world, Stathis #1 may be right. This would mean Stathis first assumption was a first person assumption, but the whole point of Stathis seems (to me) third person. Also what would be the meaning of physical in a first person assertion. Perhaps Stathis could comment. Now you are right we should agree on what we deem POSSIBLE. With the comp hyp I argued that POSSIBLE = arithmetically consistent, and then we can go back asking G and G* Giving that logic is not so well known apparently I will soon or later invite you all to Smullyan's knight knaves Island. It is the gentlest path to G and G* which are the propositional psychologies from which UDA shows how to extract the quantum measure in case (comp is true). And from which I have extract some bits of von neuman's quantum logic (but I am just beginning opening a vast and heavy doors here). Why not now? The native of that Island are all either knight or knaves and knight always tell the truth, and knaves always lie. You go there. Problem 1. A native tell you I am a knight. Is it possible to deduce the native's type? Problem 2. You meet someone on the island, and he tells you I am a knave. What can you deduce? I would be please to get answers, or critics. I think it will be useful if only by John Mike remark: we will not progress if we make not clear the word possible in our everything context ... Logic can help because it is the science of proofS, truthS, and possibilitieS (note the s). Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
Dear Bruno, perhaps the list will forgive me a bit of distraction upon your knight and knave koan. I call it a koan, because within your conditions there is no right solution to either of the questions. IMO Problem #1 is open, #2 is subject to unlisted circumstances. (Common sense). To make the question subject to 'common sense' logic, I put some restrictions on (my) Problem JM: In the desert watch-tower at the fork there are two guards, twins. One tells ALWAYS the truth, the other ALWAYS lies. One way leads to Bagdad, the other to the lion-desert. You can ask ONE question: to decide which way to go to Bagdad. Which one is that ONE question getting you the right answer without knowing Which brother is on duty? I give you a day, if nobody does so,tomorrow I will post the answer. John Mikes - Original Message - From: Bruno Marchal [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, July 20, 2004 12:43 PM Subject: ... cosmology? KNIGHT KNAVE At 09:55 20/07/04 -0400, John Mikes wrote: It all depends what do we deem: POSSIBLE. According to what conditions, belief, circumstances? If we accept the here and now as the world, Stathis #1 may be right. This would mean Stathis first assumption was a first person assumption, but the whole point of Stathis seems (to me) third person. Also what would be the meaning of physical in a first person assertion. Perhaps Stathis could comment. Now you are right we should agree on what we deem POSSIBLE. With the comp hyp I argued that POSSIBLE = arithmetically consistent, and then we can go back asking G and G* Giving that logic is not so well known apparently I will soon or later invite you all to Smullyan's knight knaves Island. It is the gentlest path to G and G* which are the propositional psychologies from which UDA shows how to extract the quantum measure in case (comp is true). And from which I have extract some bits of von neuman's quantum logic (but I am just beginning opening a vast and heavy doors here). Why not now? The native of that Island are all either knight or knaves and knight always tell the truth, and knaves always lie. You go there. Problem 1. A native tell you I am a knight. Is it possible to deduce the native's type? Problem 2. You meet someone on the island, and he tells you I am a knave. What can you deduce? I would be please to get answers, or critics. I think it will be useful if only by John Mike remark: we will not progress if we make not clear the word possible in our everything context ... Logic can help because it is the science of proofS, truthS, and possibilitieS (note the s). Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT KNAVE
What would your other brother say is the road to Baghdad? Then take the other direction! Cheers On Tue, Jul 20, 2004 at 06:18:43PM -0400, John M wrote: Dear Bruno, perhaps the list will forgive me a bit of distraction upon your knight and knave koan. I call it a koan, because within your conditions there is no right solution to either of the questions. IMO Problem #1 is open, #2 is subject to unlisted circumstances. (Common sense). To make the question subject to 'common sense' logic, I put some restrictions on (my) Problem JM: In the desert watch-tower at the fork there are two guards, twins. One tells ALWAYS the truth, the other ALWAYS lies. One way leads to Bagdad, the other to the lion-desert. You can ask ONE question: to decide which way to go to Bagdad. Which one is that ONE question getting you the right answer without knowing Which brother is on duty? I give you a day, if nobody does so,tomorrow I will post the answer. John Mikes -- *PS: A number of people ask me about the attachment to my email, which is of type application/pgp-signature. Don't worry, it is not a virus. It is an electronic signature, that may be used to verify this email came from me if you have PGP or GPG installed. Otherwise, you may safely ignore this attachment. A/Prof Russell Standish Director High Performance Computing Support Unit, Phone 9385 6967, 8308 3119 (mobile) UNSW SYDNEY 2052 Fax 9385 6965, 0425 253119 () Australia[EMAIL PROTECTED] Room 2075, Red Centrehttp://parallel.hpc.unsw.edu.au/rks International prefix +612, Interstate prefix 02 pgpBWHQowHwOp.pgp Description: PGP signature
Re: ... cosmology? KNIGHT KNAVE
Bruno, John, Russell I am half-way through Smullyan's book. It is an entertaining book for someone motivated enough to do all these puzzles, but I think that what is missing is a metalevel discussion of what all this means. Mathematical fireworks occur because we are dealing with self-referential systems. In the old days they may have called them reflexive. Reflection is, I think, an essential component of conscious thought. The type of reflection I have encountered so far in the book involves infinite reflections which lead to paradoxes. For example if someone says I am a knave then obviously we have a paradox. The human mind, however, does not have the capacity to deal with an infinite number of reflections. (I think that you think that I think that you think.). If, however, self referential systems are limited to a finite number of reflections, such as the human mind is capable of, then these paradoxes may go away. With one reflection a knave would says: I am a knight. With two reflections he would say I am a knave. With three reflections, I am a knight. With four I am a knave. and so on. With an infinite number of reflections he would remain Forever Undecided. I am not sure if Physics is derived from an ideal infinite self-referential systems or from a more human and messy finite system and I cannot think of an obvious and clear-cut justification for either approach. What do you think? George Bruno Marchal wrote: At 09:55 20/07/04 -0400, John Mikes wrote: It all depends what do we deem: POSSIBLE. According to what conditions, belief, circumstances? If we accept the here and now as the world, Stathis #1 may be right. This would mean Stathis first assumption was a first person assumption, but the whole point of Stathis seems (to me) third person. Also what would be the meaning of physical in a first person assertion. Perhaps Stathis could comment. Now you are right we should agree on what we deem POSSIBLE. With the comp hyp I argued that POSSIBLE = arithmetically consistent, and then we can go back asking G and G* Giving that logic is not so well known apparently I will soon or later invite you all to Smullyan's knight knaves Island. It is the gentlest path to G and G* which are the propositional psychologies from which UDA shows how to extract the quantum measure in case (comp is true). And from which I have extract some bits of von neuman's quantum logic (but I am just beginning opening a vast and heavy doors here). Why not now? The native of that Island are all either knight or knaves and knight always tell the truth, and knaves always lie. You go there. Problem 1. A native tell you I am a knight. Is it possible to deduce the native's type? Problem 2. You meet someone on the island, and he tells you I am a knave. What can you deduce? I would be please to get answers, or critics. I think it will be useful if only by John Mike remark: we will not progress if we make not clear the word possible in our everything context ... Logic can help because it is the science of proofS, truthS, and possibilitieS (note the s). Bruno http://iridia.ulb.ac.be/~marchal/
