Fitch's paradox and OWA
in [Logic]
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From: Nam Nguyen on 31 Dec 2009 23:44 Daryl McCullough wrote: > Barb Knox says... >> In article >> <a3f061ed-3838-4be9-b73a-836141dc640f(a)u7g2000yqm.googlegroups.com>, >> Marshall <marshall.spight(a)gmail.com> wrote: > >>> I was more under the impression that Goedel showed there >>> was no complete finite theory of them, rather than no >>> way to define them. Are you saying those are equivalent? >> Yes, in this context. Since we are finite beings we need to use finite >> systems. > > I don't agree. What Godel's theorem says is that we can't know all > the truths about the natural numbers, but it doesn't imply that there > is any fuzziness in what we mean by natural numbers. For what it's worth, in mathematics the truths about the natural numbers is what we mean we mean by the natural numbers, not what they existentially are! > > All the nonstandard models of the naturals contain infinite objects. > We're not likely to mistake such an object for an actual natural. As > you say, we are finite beings, so any natural we can write down is > finite. But there's no such thing as finite natural numbers. Natural numbers are natural numbers, infinite or not.
From: Jan Hidders on 1 Jan 2010 04:51 On 1 jan, 05:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Marshall says... > > >However what I was referring to was specifically > >how they get from step 7 to step 8 within that > >RAA proof. Your response does not seem to > >address that particular issue. > > That's exactly the step that I was talking about. > Steps (4), (5) and (6) and (7) constitute a proof > of ~K(p  ⧠~Kp). Therefore, we have > |- ~K(p  ⧠~Kp) > > By C, if you have |- f, then you have |- [] f. > Letting f = ~K(p  ⧠~Kp), it follows that > |- [] ~K(p  ⧠~Kp) > which is step (8). > > >Are your comfortable with how step 8 is > >obtained from step 7 via Rule C as described > >on this page? > >http://plato.stanford.edu/entries/fitch-paradox/ > > Yes, that's exactly what they are doing. They > didn't use the |- symbol in step 7, but it is > clear that (7) is the conclusion of a proof. Wow. You are right. They correctly conclude in (7) that |- ~K(p  ⧠~Kp). Hmm. I need to think this over. I'm beginning to believe now that the inference in the paradox is in fact correct. -- Jan Hidders
From: Jan Hidders on 1 Jan 2010 08:07 On 31 dec 2009, 18:47, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Okay, I've thought about it a little more, and I have come to > the conclusion that Fitch's paradox is invalid. Or perhaps the > statement of the knowability principle is wrong. > > Here's the proof of the contradiction: > > 1. (Knowability principle) For all p: p -> <> K(p) > > where <>Phi means "Phi is possibly true" and K(Phi) means > "Phi is known". > > 2. (Non-omniscience principle) For some p: p & ~K(p) > > 3. Letting p0 be the true but unknown proposition, we have > p0 & ~K(p0) > > 4. From 1&3, we have <>K(p0 & ~K(p0)) > > At this point, let me switch to possible world semantics: <> Phi > means "Phi is true in some world". So let's switch to the world > in which K(p0 & ~K(p0)) is true. In that world we have > > 5. K(p0 & ~K(p0)) > > From this it follows: > > 6. K(p0) & K(~K(p0)) > > But only true things are knowable, so from K(~K(p0)) it > follows that ~K(p0). So we have > > 7. K(p0) & ~K(p0) > > which is a contradiction. > > The mistake becomes clearer if we explicitly introduce > possible worlds. Let's use w ||- Phi to mean "Phi is true > in world w" and K_w(Phi) to mean "Phi is known in world > w". Let's introduce w0 to mean "our world". Then the > proof becomes the following: > > 1. (Knowability principle) for all p: (w0 ||- p) -> exists w, K_w(p) > > In other words, if p is true in our world, then there exists another > world in which p is knowable. > > 2. (Non-omniscience principle) for some p: w0 ||- p & ~K_w0(p) > > 3. Introducing the constant p0 for this unknown proposition, we > have: w0 ||- p0 & ~K_w0(p0) > > 4. From 1&3, we have exists w, K_w(p0 & ~K_w0(p0)) > > 5. Letting w' be a name for some world making the existential true, > we have: K_w'(p0 & ~K_w0(p0)) > > From this it follows: > > 6. K_w'(p0) & K_w'(~K_w0(p0)) > > Since only true things are knowable, we have: > > 7. K_w'(p0) & ~K_w0(p0) > > That's no contradiction at all! The proposition p0 is > known in one world, w', but not in another world, w0. > It only becomes a contradiction when you erase the > world suffixes. True, but you have now fundamentally changed the semantics of the K operator in the sense that the model theory now looks very different. You have essentially turned K from a unary operator K(p) to a binary operator K(w,p). If you assume the model theory that I presented earlier the inferences can be verified to be in fact all correct (with apologies for copying your words): Let W be the set of all possible worlds, and w0 the element of W that is our world in W. Since W will be fixed I will omit it in (W, w) ||- Phi and simply write w ||- Phi. 1. (Knowability principle) for all p : (w0 ||- p) -> exists w in W, w ||- K(p) 2. (Non-omniscience principle) For some p: w0 ||- p & ~K(p) 3. Introducing the constant p0 for this unknown proposition, we have: w0 ||- p0 & ~K(p0) 4. From 1&3, we have: Exists w in W, w ||- K(p0 & ~K(p0)) 5. Letting w' be a name for some world making the existential true, we have: w' ||- K(p0 & ~K(p0)) From this by principle (A) it follows: 6. w' ||- K(p0) and w' ||- K(~K(p0)) Since only true things are knowable by principle (B), we have: 7. w' ||- K(p0) and w' ||- ~K(p0) From the semantics of K in the model theory it then follows that: 8. K(p0) in w' and K(p0) not in w' Which is indeed a contradiction. -- Jan Hidders
From: Marshall on 1 Jan 2010 10:28 On Dec 31 2009, 7:24 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Marshall wrote: > > Can you verify that your definition of the naturals meet the definition of > formal system model, with say Q is the underlying system at hand, as I did > verify M w.r.t to T above? [It's just a pure simple technical question!] I expect so. I haven't gone through the exercise at length, but the whole process seems straightforward enough. I have done certain individual proofs, but not all of them. > > What difficulties do you foresee? > > Ok. this is a much better and more technical question one could entertain.. > > In a nutshell, one of difficulties that formula such as (1) or (1') presents is > that there's no way you could define any model of Q such that a certain expected > set of 2-tuples (i.e. _relation_) can be verified to exist. And if you can't, > you can't tell whether or not you have would conform to the overall definition of a > model of the underlying formal system (say Q in this case). If you don't admit the existence of any possible technique of showing such a thing, it wouldn't be a good use of my time to try to convince you otherwise. Note at least that yours is a minority opinion, here; it's certainly possible to show the necessary relations exist, and have the desired properties, though such ordinary methods as induction. Marshall
From: Daryl McCullough on 1 Jan 2010 10:28
Jan Hidders says... > >On 31 dec 2009, 18:47, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> From this it follows: >> >> 6. K_w'(p0) & K_w'(~K_w0(p0)) >> >> Since only true things are knowable, we have: >> >> 7. K_w'(p0) & ~K_w0(p0) >> >> That's no contradiction at all! The proposition p0 is >> known in one world, w', but not in another world, w0. >> It only becomes a contradiction when you erase the >> world suffixes. > >True, but you have now fundamentally changed the semantics of the K >operator in the sense that the model theory now looks very different. >You have essentially turned K from a unary operator K(p) to a binary >operator K(w,p). That's not a change of the *semantics*. That's a change of the *syntax*. My claim is that in the possible worlds semantics, every predicate (and operator) that can vary from world to world implicitly is a function of the world. That complexity can usually be avoided because implicitly we assume that everything is talking the same world. But when you nest <> and K, it is no longer possible to make that assumption. Not without restrictions on what can be said. My point is that the knowability principle doesn't make any sense without explicit mention of possible worlds. It might make sense if we restrict the principle to propositions p that don't involve the knowability operator. But if we restrict it that way, we can't carry out Fitch's proof. -- Daryl McCullough Ithaca, NY |