The End of an Era - That of The Natural Numbers
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From: Charlie-Boo on 18 May 2010 14:38 On Apr 21, 11:50 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > In this post, we'll demonstrate > that if there's an intuition of knowing the natural numbers, > there's also an intuition about not knowing them that > would invalidate GIT proof. Ok, then how about this one: The set of provable sentences is recursively enumerable but the set of true sentences is not, so truth does not coincide with provability? Is that invalidated, too? Or even better: "This is not true." has no truth value. "This is not provable" has a truth value, viz. true. So again true does not equal provable. Are my two proof of Godel's 1st. Incompleteness Theorem also invalidated? (If they are, I can show you a bunch more - just see my past posts, actually.) C-B > Let F and F' be 2 formulas, we'll define the logical operator > xor as: > > F xor F' <-> (F \/ F') /\ ~(F /\ F') > > Let's now consider thew 2 formulas in L(PA): pGC ("pro GC") > and cGC("counter GC"): > > pGC <-> "There are infinitely many examples of GC" > cGC <-> "There are infinitely many counter examples of GC" > > Now, let's consider the following formula: > > (1) pGC xor cGC > > First observation: if we have any intuition about the naturals > then we'd also have the intuition that we can't know the arithmetic > truth or falsehood of (1). > > Second observation: as a consequent of the 1st observation, no > extension of Q can prove nor disprove (1) unless it's inconsistent. > > Third observation, as a consequence of the 2nd observation, _if_ a > formal system T is consistent and capable of expressing arithmetic > then (1) is undecidable in T. Hence G(T) is no longer necessary in > T's incompleteness, again, _if_ it is indeed incompleteness. > > Finally, as a consequence of all the above, if T is capable of > expressing arithmetic of the naturals then it's impossible to > demonstrate (prove) T's incompleteness. Hence GIT is invalid.
From: Nam Nguyen on 18 May 2010 23:05 Alan Smaill wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Alan Smaill wrote: >>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>> >>>> Nam Nguyen wrote: >>>>> Marshall wrote: >>>>>> By your own definition of "true in a theory", this is not >>>>>> the question. Your definition was "true in all models of >>>>>> a consistent theory." Since T is not consistent, the >>>>>> meaning (by *your* definition) of "true in T" is undefined. >>>>> We live in a binary logic world remember? By _default_, if a formula >>>>> doesn't meet the definition of being true in a context then it is >>>>> defined to be false in that context. In technical terms, which >>>>> I did mention in a conversation with Aatu before, a formula F is >>>>> true in PA: >>>>> >>>>> > F is true <-> (PA isn't inconsistent) and (PA |- F) >>>>> >>>>> Just replace PA by my specific inconsistent T and note my >>>>> "and" above. Do you see now x=x is false in T? >>>> Btw, that's not my definition as you mentioned above. Although he >>>> used a slightly different word "valid", the definition could be >>>> found in Shoenfield's book: >>>> >>>> "A formula is valid in T if it is valid in every model of T" >>>> (Pg. 22) >>>> >>>> Note also what I had stipulated above with Aatu and what I said to >>>> Alan are equivalent. >>> It looks like you are claiming Shoenfield's formulation >>> and yours are *equivalent* (after replacing "valid" with "true"). >> On pg. 18 (on a "Structure", say, M) he had: >> >> "We want to define a formula A to be valid in M if all the meanings >> of A are true in M". >> >> So he defined being "valid" as being "true". > > let's leave the terminology aside, and look at the logic. > >>> They are not: Shoenfield's version allows a formula to be valid even >>> for an inconsistent T, and yours does not. >> Where did he assert or stipulate that? > > When he said: > > "A formula is valid in T if it is valid in every model of T" How does that invalidate a formula is being false in an inconsistent T? > > How would you express this in FOL? What do you mean by "this" here?
From: William Hughes on 18 May 2010 23:39 On May 19, 12:05 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Alan Smaill wrote: > > When he said: > > > "A formula is valid in T if it is valid in every model of T" > > How does that invalidate a formula is being false in an inconsistent > T? > Let F be the formula and T be an inconsistent theory. Then F is valid in every model of T (vacuously true) Therefore F is valid in T. Your interpretation is 'he defined being "valid" as being "true"'. Therefore F is true in T. You demand that F must be either true or false in T. Therefore F is not false in T. - William Hughes
From: Nam Nguyen on 19 May 2010 01:26 William Hughes wrote: > On May 19, 12:05 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> Alan Smaill wrote: > >>> When he said: >>> "A formula is valid in T if it is valid in every model of T" >> How does that invalidate a formula is being false in an inconsistent >> T? >> > > Let F be the formula and T be an inconsistent theory. > > Then F is valid in every model of T (vacuously true) No. FOL formula-truth definition however you define it would conform to Tarski's concept of truth which doesn't harbor the concept of "vacuous truth" and which is NOT of inference nature. If the definition of being true is not met then being false _automatically_ is assumed, by definition. If an F's being true in a T requires T be consistent (i.e. having models) and T is inconsistent (i.e. having no true models) then F is false in T, by default definition of false. In the current FOL, "F is valid in every model of T" is equivalent to "T is consistent _and_ F is provable". So, given T = {(x=x) /\ ~(x=x)} is inconsistent, then F is invalid in T. Now just replace "valid" by "true" and "invalid" by "false", the F is false in this T. That's by definition, by meeting the definition requirements - not by inference! > > Therefore F is valid in T. > > Your interpretation is 'he defined being "valid" as being "true"'. > Therefore F is true in T. You demand that F must be either > true or false in T. Therefore F is not false in T. I've refuted your argument (above). In summary, if the conditions for meeting the definition of being true exist then being true is the case, otherwise being false will be the case. There's nothing in between - not even the inferred "vacous truth".
From: Nam Nguyen on 19 May 2010 01:45
Nam Nguyen wrote: > William Hughes wrote: >> On May 19, 12:05 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>> Alan Smaill wrote: >> >>>> When he said: >>>> "A formula is valid in T if it is valid in every model of T" >>> How does that invalidate a formula is being false in an inconsistent >>> T? >>> >> >> Let F be the formula and T be an inconsistent theory. >> >> Then F is valid in every model of T (vacuously true) > > No. FOL formula-truth definition however you define it would conform > to Tarski's concept of truth which doesn't harbor the concept of > "vacuous truth" and which is NOT of inference nature. If the > definition of being true is not met then being false _automatically_ > is assumed, by definition. > > If an F's being true in a T requires T be consistent (i.e. having > models) and T is inconsistent (i.e. having no true models) then F is > false in T, by default definition of false. > > In the current FOL, > > "F is valid in every model of T" > > is equivalent to > > "T is consistent _and_ F is provable". > > So, given T = {(x=x) /\ ~(x=x)} is inconsistent, then F is invalid > in T. Now just replace "valid" by "true" and "invalid" by "false", > the F is false in this T. > > That's by definition, by meeting the definition requirements - not > by inference! > >> >> Therefore F is valid in T. >> >> Your interpretation is 'he defined being "valid" as being "true"'. >> Therefore F is true in T. You demand that F must be either >> true or false in T. Therefore F is not false in T. > > I've refuted your argument (above). In summary, if the conditions > for meeting the definition of being true exist then being true is > the case, otherwise being false will be the case. There's nothing > in between - not even the inferred "vacous truth". In other words, given the statement: (*) If the Moon is made of cheese then the Moon is blue (*) as a whole statement is a true statement by the inference nature of definition of (*). And you could say, as a manner of speaking, "the Moon is blue" is vacuously true, per this inferential statement. But the Moon is factually NOT blue and hence can't be blue on the account of "the Moon is made of cheese", or other inferential account for that matter. Do you understand the difference between being "vacuously true" and being true? |