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
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 -> ] 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 some
Re: ... cosmology? KNIGHT & KNAVE
Tell me again where I am going wrong. 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? 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. 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. Therefore it is true that Bp->Bq. 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. 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. 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. When the machine writes 117.q, it doesn't mean anything to the machine. To us it means that the machine believes q, or that the machine believes q is true. Given this approach, I am very hesistant to say that 191. Bp means that the machine believes that it believes p. I have no problem saying that it means that the machine believes Bp. But to say that the machine "believes that it believes p" uses the word "believes" in two very different and confusing ways. The first "believes" is just a statement about what the machine has derived as a theorem. We choose to say that the machine's theorems are what it "believes". I am OK with that. But the second "believes" refers to the letter B, which we are arbitrarily choosing to identify with this word "belief". To the machine, B is just a letter. I still say that I need to know what the rules are that the machine will apply to that letter. I see that I was wrong to think that p -> Bp was a rule the machine would have if it were "normal". You said that for a normal machine, if it ever proved p, it would also prove Bp. Okay, but how could it possibly do this without ANY rules to deal with the letter B? Normality is not something one can just assert about a machine. You would need to give it a rule for dealing with the letter B, then you could prove (not the machine proving, you would prove it) that if the machine ever derived p, its rules for dealing with the letter B would then cause it to derive Bp. In this way you would show that the machine was normal. My axiom, which I should have written as, "0a. for all x, x->Bx", would in fact be sufficient to show that a machine which had that axiom would be normal. A machine which had this rule for dealing with the letter B would be normal, because any time it derived p, it could immediately derive Bp using this axiom. However, this machine may be too powerful. Although it is normal, it is much more. So my question to you is, what is an example of an axiom for dealing with the letter B that would make a machine be "just" normal, but no more? Hal Finney
Re: ... cosmology? KNIGHT & KNAVE
Hal, I didn't see the whole post. So here is a sequel. At 11:01 28/07/04 -0700, Hal Finney wrote: BM: > 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. HAL: 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. No, I stop you right here. You redo the error. No problem, I still do it 1 or 2 times a day. And, would have I not made it persistently in my youth, I would be much farther. You must be very careful reading a proposition in english, and then translating in the machine language. Your axiom 0a says that the truth of x implies the truth of Bx. It is really the usual implication: x implies Bx 1 1 1 1 0 0 0 1 1 0 1 0 Would it be true. Then would x be true, Bx would be true too. "x implies Bx" means either x is false or Bx is true. It really means that the machine is truth complete, that is omniscient, that is ... the contrary of the consequence of Godel theorem. It is also the reason why i use generally "->" instead of "implies". Ah, perhaps that is the root of the confusion. Same remark for your axiom 0b. Bx implies x, or better, Bx -> x, means that the machine is accurate. It is built in such a way that if Bx is true (that is if the machine believe x, then x is true). We can sum up: 0a = the machine believes all the truth. 0b = the machine believes only the truth. (And so new exercise: what happens with complete and /or accurate machine of type 1 when they meet the Knight saying the usual proposition?) Normality and stability are much weaker notion: A machine is normal means: if the machine ever believe p then sooner or later the machine will believe Bp. Now, please, just translate *carefully* this in the language of the machine. Let us do it by stage: the machine believe p is Bp the machine believe Bpis BBp if the machine believes p then the machine believes Bp becomes Bp implies BBp, or perhaps better: Bp -> BBp The same for stability, it will gives BBp -> Bp. Summing up: Stability is: BBp -> Bp Normality is: Bp -> BBp I don't find these words to be particularly appropriate, by the way, but I suppose they are traditional. More or less traditionnal. Normality is very traditionnal, although in slightly strict or broader sense. Stability is less tarditionnal, but Smullyan uses that term. Anyway, it is not logically relevant ... 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). Be reassured. Our reasoner believes in the excluded middle: that is they believe in (x v -x). In machine Lang.: B(x v -x), but this will not imply (Bp or B~p) Unless your axiom 0a is true, but that will be impossible, by Godel. No machine can be complete (except inconsistent one). I will read the rest at ease, but I think it follows from your axioms 0a, and 0b, but 0a does not apply for any reasoner who believes in arithmetic ... The rest of your post is correct deduction, but with wrong premise. When you said: 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. Yes but the machine believes q is true, or assert q is true. The machine is not asserting it believes true. 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 machine's numbered list, which it is, right there. The machine believes Bp, but Bp is not (necessarily) true. BBp is necessarily true. Another example 207. p implies q By the same reasoning, this does not mean that if p is true, then q is true. It means that if the machine ever believes p (i.e. it writes it down in its list of theorems), then it will believe q (i.e. it will eventually write down q in its list of theorems). The true statment is that Bp implies Bq. You can also (trivially
Re: ... cosmology? KNIGHT & KNAVE
At 11:01 28/07/04 -0700, Hal Finney wrote: 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. OK. Let us call it the *intended meaning* of "I believe p" or of "Bp". The other meaning is the string "Bp". Well, *this* is confusing! I would prefer to avoid such move. You could have said that the meaning of "I believe p" is the string contitutes by "I", "blank", "b", "e", ... In which case "je crois p" (I believe p, in french) has not the same meaning as "I believe p". But it is preferable that "meaning" is defined in a way which makes it independent of the chosen language. 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". But we do say that "belachen" and "believe" have the same meaning(*) ... The question of who attributes the meaning is of course a hard philosophical question which leads eventually to the qualia problem. Some "neurophilosopher" say that consciousness does not exist, there is only firing neurons ... You can always believe that humans or machine are zombie ... Actually the path I propose will explain why machines themselves will be led to similar remarks ..., and will explain why we cannot really hope for much more. Also "Bp" will be translated later in arithmetic at some point, and this will make clear its meaning at the arithmetical level. All I need to know, as a formal system, is what rules the letter B follows. Even here a lot of philosophers will say that such rules will not give the meaning. I really think this debate is a little premature. After all you did solve the puzzle, and you did find which rule B should follow for extracting something non trivial from the machine encounter with the knight. (I guess you did grasp the correction, isn't it?). To sum up: Bp, Ich glaub p, Io credo p, I believe p, je crois p, etc. have all the same intended meaning, which is the one you gave, that you (as a machine) will print soon or later p. Now a machine could print Bp and never print p, in which case we will say the machine is not really accurate, a little like a computer saying that there is a stack overflow when actually there is no stack overflow. OK? We will tackle the question "does the machine have a genuine conscious meaningful qualia-full experience of what she believes?" later, for that is an important but very hard and premature question imo. Bruno (*) Are you sure of "belachen" for "believe" in German? Isn't it "glaube"? http://iridia.ulb.ac.be/~marchal/
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,
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 exis
Re: ... cosmology? KNIGHT & KNAVE
Hi John, At 17:19 26/07/04 -0400, John M wrote: Bruno, (and Class) 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
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
Bruno, (and Class) We have an overwhelming ignorance about Ks and Ks. We don't know their logical built, their knowledege-base, their behavior. 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). 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? Are they robots or humans? Looks like machines. Are machines omniscient? To your present 2 problems: none of them CAN be a knight, because in all fairness, nobody knows what may a person 'know' or 'believe' in the future. If I go to the Office of Records, I may learn what that 'K' is. John Mikes - Original Message - From: "Bruno Marchal" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Monday, July 26, 2004 12:49 PM Subject: Re: ... cosmology? KNIGHT & KNAVE > Hi, > > > At 19:47 23/07/04 +0200, I wrote: > >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" ? > > > It was perhaps not pedagogical to say "big" and "very big". > Here John Mikes would be accurate to say those are not problems, > but koans. So you can *meditate* on it ... > My intent in the use of the word "big" was more to point on the fact > that ,going from the preceding problems to those new one is, > as you can guess, the passage from ordinary logic > to modal logic, giving that "knowing" and "believing" are modalities. > Still we do have intuition on what knowledge and believe can be, and > we can try to get some conclusion. > > Let us try the problem 5. Just to be sure let me know if you agree > that the proposition: > > You will not know that Lance Amstrong has win the "Tour de France", > is false. I mean the negation of "You will not know p" is "you will know p". > > OK then. Suppose the native is a knave. Then it means he is lying when > he say that "you will never know I am knight". But that means I will know > he is a knight. But I cannot *know* he is a knight if he is a knave, so he > cannot be > a knave, and thus he is a knight. But having reach that conclusion I know > he is a knight, but then he was lying (giving he told me I will never > know that. So he is a knave, yes but we have already show he cannot > be a knave. It looks like we are again in an oscillating state of mind. > Is it a tourist again? No. I told you it is a native and all native are, > by the definition of the KK island, either knight or knave. So what? > > I will give many critics later on some "less good" chapter of FU, (ex: > the chapter "the heart of the matter" is too short, or when he identify > a reasoner with a world in his possible world chapter, etc.) > But Smullyan's discussion on the problem 5 is really quite instructive. > Its attack of the problem 5 is quite alike my attack on the mind body > problem, Smullyan will interview "reasoner" (actually machine) on those > questions. And he goes on very meticulously by defining a hierarchy > of type of reasoner, making those problem, mainly the problem 6 more > and more interesting. > Indeed, with the problem 5, we are quickly done, giving that the > problem 5 leads toward a genuine contradiction even with the reasoner > of lowest type: the one Smullyan calls the "reasoner of type 1". > > A reasoner, or a machine, or a system (whatever) will be said to be > of type 1 if by definition the following conditions hold: > > 1) he believes all classical tautologies (our lowest type of reasoner > is already a platonist!) > 2) if the reasoner ever believes both X and X ->Y, he will believe Y. > > I will also say, following Smullyan and ... Theaetetus, that a reasoner > know p, in the case the reasoner believes p and p is true, that is when > the reasoner correctly believes p. > There is nothing metaphysical in our use of the word "believe". You can > substitute it by "prove" or even just "print". That would mean you are > in front of a sort of theorem prover which will, for any classical tautology, > print it one day or an other (condition 1), and, in case it prints X and X->Y, > it will, soon or later, print Y. That is he prints a proposition if and > only if he believes it. > > To transform the "koan 5" into a genuine problem, I must explain what it > means for a reasoner to believe in the rules of the island. It means that > if he met a native asserting a pro
Re: ... cosmology? KNIGHT & KNAVE
Hi, At 19:47 23/07/04 +0200, I wrote: 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" ? It was perhaps not pedagogical to say "big" and "very big". Here John Mikes would be accurate to say those are not problems, but koans. So you can *meditate* on it ... My intent in the use of the word "big" was more to point on the fact that ,going from the preceding problems to those new one is, as you can guess, the passage from ordinary logic to modal logic, giving that "knowing" and "believing" are modalities. Still we do have intuition on what knowledge and believe can be, and we can try to get some conclusion. Let us try the problem 5. Just to be sure let me know if you agree that the proposition: You will not know that Lance Amstrong has win the "Tour de France", is false. I mean the negation of "You will not know p" is "you will know p". OK then. Suppose the native is a knave. Then it means he is lying when he say that "you will never know I am knight". But that means I will know he is a knight. But I cannot *know* he is a knight if he is a knave, so he cannot be a knave, and thus he is a knight. But having reach that conclusion I know he is a knight, but then he was lying (giving he told me I will never know that. So he is a knave, yes but we have already show he cannot be a knave. It looks like we are again in an oscillating state of mind. Is it a tourist again? No. I told you it is a native and all native are, by the definition of the KK island, either knight or knave. So what? I will give many critics later on some "less good" chapter of FU, (ex: the chapter "the heart of the matter" is too short, or when he identify a reasoner with a world in his possible world chapter, etc.) But Smullyan's discussion on the problem 5 is really quite instructive. Its attack of the problem 5 is quite alike my attack on the mind body problem, Smullyan will interview "reasoner" (actually machine) on those questions. And he goes on very meticulously by defining a hierarchy of type of reasoner, making those problem, mainly the problem 6 more and more interesting. Indeed, with the problem 5, we are quickly done, giving that the problem 5 leads toward a genuine contradiction even with the reasoner of lowest type: the one Smullyan calls the "reasoner of type 1". A reasoner, or a machine, or a system (whatever) will be said to be of type 1 if by definition the following conditions hold: 1) he believes all classical tautologies (our lowest type of reasoner is already a platonist!) 2) if the reasoner ever believes both X and X ->Y, he will believe Y. I will also say, following Smullyan and ... Theaetetus, that a reasoner know p, in the case the reasoner believes p and p is true, that is when the reasoner correctly believes p. There is nothing metaphysical in our use of the word "believe". You can substitute it by "prove" or even just "print". That would mean you are in front of a sort of theorem prover which will, for any classical tautology, print it one day or an other (condition 1), and, in case it prints X and X->Y, it will, soon or later, print Y. That is he prints a proposition if and only if he believes it. To transform the "koan 5" into a genuine problem, I must explain what it means for a reasoner to believe in the rules of the island. It means that if he met a native asserting a proposition p, then he believes the proposition "the native is a knight if and only if p is true". We will write Bp for the reasoner believes p, and Kp for the reasoner knows p. For exemple we have Kp <-> p & Bp problem 5' (ameliorated version): A visitor of type 1 meets, on the KK island, a native telling him "you will never know I am a knight". Convince yourself that this really cannot happen. Derive a contradiction. Problem 6' (new version): A visitor of type 1 meets, on the KK island, a native telling him "you will never believe I am a knight". Convince yourself that this can happen. Derive as many conclusions as you can (note that all FU will follow, and even my thesis!!! (if you are patient enough). The idea is to give more and more self-awareness to the visitor ... Curiously enough that path converges. Bruno PS Those who feels overwhelmed can wait I come back on problem 4, when I will recall a little bit more matter on propositional logic. Apology for my post to Jesse which could look a little bit "advanced" for many (it presupposes FU and a little bit more). Actually writing this stuff helps me for my paper. I thank you in advance. http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT & KNAVE & beyond.
At 16:58 23/07/04 -0400, Jesse Mazer wrote: Bruno Marchal wrote: All right. But modal logic are (traditionaly) extension of classical logic, so that causal implication, or >natural language entailment, when study mathematically are generally defined through modalities + >"material implication". So in a sense, you confuse yourself by premature anticipation. Well, I guess "in every possible world where X is true, Y is true also" can only be false if there's a possible world where the classical logical statement "X -> Y" is false (because in that possible world, X is true but Y is false). So perhaps the possible-world statement would be equivalent to the modal-logic statement "it is necessarily true that X->Y"--would this be an example of modal logics "extending" classical logic? Exactly. In any case, in classical logic X -> Y can only be false if X is true in *our* world, I would say: in our arithmetical platonia (see below). whereas the possible-world version of "if X then Y" does not require that X is true in our world, although it must be true in some possible world. X could be false in all possible world. "if X then Y" will then be true in all possible worlds. And like I said, I think the possible-world statement more accurately captures the meaning of the natural-language statement. Yes but it is here that you anticipate, relatively to the goal which we have ascribe to ourselves. To sum up in a nutshell: the UD Argument shows comp transforms physics into a "science of" a measure on all (relative) computational histories. Those comp histories are relatively described in Arithmetical Platonia (the set of all true arithmetical propositions). In particular physics must appears in the discourse of a Self-Referentially Correct machine (SRC machine) when interviewed on the geometry of their (maximal) consistent extensions. p belongs to a consistent extension of a machine M when the machine does not believe in -p (that is -B-p is true for the machine). The SRC machine is supposed to be: 1) platonistic (the machine believes the classical tautologies) 2) arithmetical platonistic (the machine believes in the theorem of PA, say) 3) computationalist (by UDA the machine believs p -> Bp, with p atomic, that is the atomic "physical consistent state" are those generated by the DU. I will come back on this remember just that comp is translated by the modal formula p -> Bp "1)" and "2)" makes the SRC machine a Loebian machine. It makes it, in FU's terminology a consistent, stable, normal, modest reasoner of type 4. That is, a reasoner of type G. So we get a modal logic, from which we can study the corresponding "multiverse" (unlike those who want to capture directly "ordinary natural language deduction" by an ad hoc choice of a modal logic) "3)" forces us to add the comp axiom. I give the name "1" to the formula "p->Bp" (1 is a shorthand for Sigma_1). We get a new theory which is just G+1 (+and a weakening of the substitution rule: p can only represent an atomic proposition [do you see why?]). Now A. Visser, from Utrecht (in Holland) has proved the arithmetical completeness for both (G+1) and its corresponding * logic (G+1)*. In honor to Visser I call often V and V* the logic G+1 and (G+1)*. See the (big) bible for a proof. Big bible = George Boolos 1993. (this is beyond FU little bible). If you identify a logic with its set of theorems you have the following diamond where the edges represent inclusion: V* G* V(except that I'm too lazy for drawing the edges) G Going up in the north west direction is the non trivial godelian passage from provability (believability) to truth. Going up in the north east direction is the non trivial comp direction. Sometimes to fix the things I say that G gives science and G* gives theology, V gives comp-science and V* gives comp-theology. (But take this with some grain of salt). Note that G* minus G (resp. V* minus V) gives all the unbelievable (comp) but true propositions. OK. At that stage we are not yet in a position to get physics. What is missing? People on the list should be able to guess giving I insist on this all the time. What is missing is the fundamental distinction between the first and third person points of view, without which the UD Argument just doesn't start. The four G, G*, V, V* gives only 3 person descriptions. G for exemple gives a logic of self-referentially correct discourse, but the machine talk about itself only from some description made (by construction) at the right level. But the UDA shows physics appears through machine's first point of view. Also G does not work for describing a probability logic. Although the box []p in G describes p as true in all accessible worlds, there exists necessarily (by Godel seconf theorem) accessible world which are cul-de-sac worlds. We would like to have []p -> <>p, or in FU's notation Bp -> -B-p. If the proba that p is one, we w
Re: ... cosmology? KNIGHT & KNAVE
At 14:46 23/07/04 -0400, Stephen Paul King wrote: 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. Still postulating a physical universe I see ... But to justify the "physical" universe without postulating it is one of our (at least mine) goal. When you say <>, you seem to abstract away my thesis, for which FU is a preliminary. See my current post to Jesse for yet another attempt for a short description of that "effort". 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. Ok, but then (arithmetical) platonism is among the postulates of the comp theory. Nobody ever ask you to believe the theory. Bruno http://iridia.ulb.ac.be/~marchal/
Re: ... cosmology? KNIGHT & KNAVE
Bruno Marchal wrote: All right. But modal logic are (traditionaly) extension of classical logic, so that causal implication, or >natural language entailment, when study mathematically are generally defined through modalities + >"material implication". So in a sense, you confuse yourself by premature anticipation. Well, I guess "in every possible world where X is true, Y is true also" can only be false if there's a possible world where the classical logical statement "X -> Y" is false (because in that possible world, X is true but Y is false). So perhaps the possible-world statement would be equivalent to the modal-logic statement "it is necessarily true that X->Y"--would this be an example of modal logics "extending" classical logic? In any case, in classical logic X -> Y can only be false if X is true in *our* world, whereas the possible-world version of "if X then Y" does not require that X is true in our world, although it must be true in some possible world. And like I said, I think the possible-world statement more accurately captures the meaning of the natural-language statement. Jesse
Re: ... cosmology? KNIGHT & KNAVE
All right. But modal logic are (traditionaly) extension of classical logic, so that causal implication, or natural language entailment, when study mathematically are generally defined through modalities + "material implication". So in a sense, you confuse yourself by premature anticipation. I know the "material" implication needs some time to be familiarized with. Bruno At 15:45 23/07/04 -0400, Jesse Mazer wrote: 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 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
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 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
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
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
> (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
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
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. 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. I compared your version to Smullyan's. The word "believe" is missing: Paraphrasing Smullyan, I quote: "If you believe I am a knight then Santa Claus exists." I read Smullyan's solution on page 124 and 125. It sounds like circular reasoning. For the rest of the list I am retyping the passage with the appropriate editing changes in deference to Santa. Beginning of Smullyan's edited quote We let k be the proposition that the native is a knight and we let C be the proposition that Santa exists. At the outset, the reasoner believes (B) the proposition BC -> C. The reasoner reasons: "Suppose I ever believe that he is a knight. Then I'll believe what he says - I'll believe that Bk -> C. Also, if I ever believe he's a knight. I'll believe that I believe he is a knight - I'll believe Bk. And so, if I ever believe he is a knight, I'll believe both Bk and Bk -> C., hence I'll believe C. Thus, If I ever believe he is a knight, then I'll believe that Santa exists. But if I ever believe that Santa exists, then Santa exists. And so, If I ever believe he's a knight, then Santa exists. Well that's exactly what he said. He said that if I ever believe he's a knight, then Santa exists, and he was right! Hence he is a knight! At this point the reasoner believes that the native is a knight and since the reasoner is normal, he continues: "Now I believe he is a knight. I have already proved that if I believe he is a knight, then Santa exists, and since I do believe that he is a knight, Santa exists." At this point the reasoner believes that Santa exists. End of Smullyan's quote. What if the reasoner had started by supposing that he believes that the native is a knave? I am confused by Smullyan. I just see it as circular reasoning. He could have started by assuming a knave and reached a different conclusion. George
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
consistency permits relevance and non-relevance, which bifurcates into potential appliable-logics. [plural] Jamie [EMAIL PROTECTED] wrote: > > Is that the same as > "Relevant, consistent statements permit logical deduction." > Bligh > > -Original Message- > From: James N Rose <[EMAIL PROTECTED]> > Sent: Jul 22, 2004 8:26 AM > To: [EMAIL PROTECTED] > Subject: 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
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
Hi All Perhaps George Levy is right and I should explain better the strategy before proceeding. The role of the logical puzzle is to encourage you to some introspection. The puzzle are easy but at some point we will give the puzzle to the machine just for studying its "psychology". You will see John Mikes is somehow right in the sense that a puzzle or a paradox can be "understand" differently according to the reasoning capacity of the machine. And we do this because the UDA has shown that if COMP is correct then physics is (re)definable as a measure put on the consistent extensions. If you peruse Smullyan's FU you will see he talks almost everywhere on consistency, and this is indeed the key notion. We will come back soon. I am aware that the knight knave island could give the appearance that such clear-cut logic cannot be serious. As I said to George, there will be a point, where a very precise link between the KK puzzle and "machine psychology" will be done, but it's premature now. Enjoy the puzzle, and if you don't like the puzzle (but are still interested to see how physics is introspectively deducible by an universal machine) try just to understand the enunciation and the solution provided. Here are simple puzzles to familiarize you with the KK island, and to recall or test your comprehension of the classical connective. I recall that the natives of the KK island are all either knave or knight. Knaves always lies, and knight always tells the truth. Problem 3 a) A visitor to the island would like to know systematically the type of the natives. He goes to the first house and ask a man his type, and the type of his spouse. He answered: "We are both knaves". What can you deduce? b) Next house. Still a couple. He asks: "Are you both knaves?". The husband answers "At least one of us is". What can you deduce? The next problem is somehow fundamental. It contains some magic which will recur again and again, and that magic will crystallize in Lob's theorem, an unexpected and deeply counter-intuitive generalization of Godel's theorem. It is fundamental for going to G and G*. Lob is a great Deutsch logician. I think he is still alive, and there is a rumor saying he lives or lived with a leopard. I recall you that the classical implication X -> Y, is false only when X is true and Y is false. Problem 4: You have an exam today. The question looks difficult, indeed the exam question is "Does Santa Claus exist?". You are panicking because you don't remember that passage of the course. But you are a modern student so you have a mobile phone, and, after having call your friends and family without success you decide in a desperate move to call the KK island. You get a native, and asks her quickly (because it is the end of the exams) 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? 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
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.
<>
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
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/
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
>
--
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email came from me if you have PGP or GPG installed. Otherwise, you
may safely ignore this attachment.
A/Prof Russell Standish Director
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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/ >
... 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/
