what to do about the Successor axiom once it reaches 999....9999#318; Correcting Math
in [Logic]
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From: Nam Nguyen on 6 Feb 2010 13:41 Nam Nguyen wrote: > > First let's assume the following languages: > > - L1 = L1(0,S,+) > - L2 = L2(0,<,*) > - L = L(0,S,+,*,<) > > Now the steps: > > Step1: Demonstrate that given a same infinite universe U of individuals, > there are an infinite list List1 described solely in L1 and an > infinite List2 in L2 that it's impossible to determine/formulate > their _exact_ alignment against each other. > > (This is in essence the FOL counterpart UT). In this step, we'll demonstrate that List1 and List2 are the 2 versions in 2 different languages of the same kind of numbers: the even numbers, and that it'd be impossible to the formulate the identity exactness, or inexactness, of the 2 lists when comparing them against each other, existentially speaking. First, assuming there's a defined P(x) in a language, the statement: "There are infinitely many examples of P" could be written as a FOL formula. But in each of L1, L2, this formula will have to be written in a different form: the Induction form (I)P for L1, and the Anti-Induction form (AI)P for L2. In details: (I)P <-> Ax[~(S(x)=0)] /\ Ex[P(x)] /\ AxEy[P(x) -> (~(y=0) /\ P(S(x+y))] (AI)P <-> Ex[P(x)] /\ AxEy[P(x) -> (P(y) /\ (x < y))] if P(x) is meant to be "even" then we'd have: (I)even <-> Ax[~(S(x)=0)] /\ Ex[even(x)] /\ AxEy[even(x) -> (~(y=0) /\ even(S(x+y))] (AI)even <-> Ex[even(x)] /\ AxEy[even(x) -> (even(y) /\ (x < y))] But how is "even(x)" defined in L1 and L2? The long and short of it is that in L1 "even(x)" can *only be defined in meta level* through induction, while it's a *pure FOL syntactical expression* in L2! In details: In L1, it's a *meta level* definition: even(x) <-> "even(0) and if even(x) then even(x+SS(x))" There are 2 counts of induction in this *meta* definition: the 1st being the definition of a term (e.g. SS(x)), and the 2nd being the definition of "even(x)" itself. [Note the self reference of "even" on both sides of "<->". In L2, by contrast, it's a *1st order* definition: even(x) <-> (x=0) \/ Eyy'[2(y) /\ x=y*y'] 2(x) <-> Ey[1(y) /\ (y Lnext x)] 1(x) <-> (0 Lnext x) /\ Ay[y = x*y] x Lnext y <-> (x < y) /\ Az[(x < z) -> (y <= z)] There are at least a couple of counts of the impossibility to determine if all the even numbers depicted in (I)even would be identical - one for one - to those in (AI)even. In the first place, we can't have 2-way syntactical translation between the 2 formulas simply because one of them is 1st order syntactically definable, while the other is not. [I'm not sure at this moment but there might a sense (AI)even not being "arithmetical" while (I)even might be]. Secondly, in term of model, if we assume the same universe U in both cases, the 2-nary relation (set) for 'S' is a *proper subset* of that for '<', hence *existentially* it's impossible to infer the larger set from the smaller one. *** In summary, given the 2 reasons above then we have to accept the fact there's something we don't know (an unknown-ability), which is that when we compare the 2 ordered lists of evens defined in (I)even and (AI)even, it's impossible to know if the 2 lists are identical, notwithstanding that they might be isomorphic in the same universe U. There might be a general way to state the above, but in essence this is the FOL UT we should accept in this step.
From: Nam Nguyen on 6 Feb 2010 16:21 Nam Nguyen wrote: > Nam Nguyen wrote: >> >> First let's assume the following languages: >> >> - L1 = L1(0,S,+) >> - L2 = L2(0,<,*) >> - L = L(0,S,+,*,<) >> >> Now the steps: >> >> Step1: Demonstrate that given a same infinite universe U of individuals, >> there are an infinite list List1 described solely in L1 and an >> infinite List2 in L2 that it's impossible to determine/formulate >> their _exact_ alignment against each other. >> >> (This is in essence the FOL counterpart UT). > > In this step, we'll demonstrate that List1 and List2 are the 2 versions > in 2 different languages of the same kind of numbers: the even numbers, > and that it'd be impossible to the formulate the identity exactness, or > inexactness, of the 2 lists when comparing them against each other, > existentially speaking. > > First, assuming there's a defined P(x) in a language, the statement: > > "There are infinitely many examples of P" > > could be written as a FOL formula. But in each of L1, L2, this formula > will have to be written in a different form: the Induction form (I)P for > L1, and the Anti-Induction form (AI)P for L2. In details: > > (I)P <-> Ax[~(S(x)=0)] /\ Ex[P(x)] /\ AxEy[P(x) -> (~(y=0) /\ P(S(x+y))] > (AI)P <-> Ex[P(x)] /\ AxEy[P(x) -> (P(y) /\ (x < y))] > > if P(x) is meant to be "even" then we'd have: > > (I)even <-> Ax[~(S(x)=0)] /\ Ex[even(x)] /\ AxEy[even(x) -> (~(y=0) /\ > even(S(x+y))] > (AI)even <-> Ex[even(x)] /\ AxEy[even(x) -> (even(y) /\ (x < y))] > > But how is "even(x)" defined in L1 and L2? The long and short of it is that > in L1 "even(x)" can *only be defined in meta level* through induction, > while it's a *pure FOL syntactical expression* in L2! In details: > > In L1, it's a *meta level* definition: > > even(x) <-> "even(0) and if even(x) then even(x+SS(x))" > > There are 2 counts of induction in this *meta* definition: the 1st > being the definition of a term (e.g. SS(x)), and the 2nd being the > definition of "even(x)" itself. [Note the self reference of "even" > on both sides of "<->". > > In L2, by contrast, it's a *1st order* definition: > > even(x) <-> (x=0) \/ Eyy'[2(y) /\ x=y*y'] > 2(x) <-> Ey[1(y) /\ (y Lnext x)] > 1(x) <-> (0 Lnext x) /\ Ay[y = x*y] > x Lnext y <-> (x < y) /\ Az[(x < z) -> (y <= z)] > > There are at least a couple of counts of the impossibility to determine > if all the even numbers depicted in (I)even would be identical - one for > one - to those in (AI)even. > > In the first place, we can't have 2-way syntactical translation between > the 2 formulas simply because one of them is 1st order syntactically > definable, while the other is not. [I'm not sure at this moment but > there might a sense (AI)even not being "arithmetical" while (I)even > might be]. > > Secondly, in term of model, if we assume the same universe U in both > cases, the 2-nary relation (set) for 'S' is a *proper subset* of that > for '<', hence *existentially* it's impossible to infer the larger > set from the smaller one. > > *** > > In summary, given the 2 reasons above then we have to accept the fact > there's something we don't know (an unknown-ability), which is that when > we compare the 2 ordered lists of evens defined in (I)even and (AI)even, > it's impossible to know if the 2 lists are identical, notwithstanding > that they might be isomorphic in the same universe U. > > There might be a general way to state the above, but in essence this > is the FOL UT we should accept in this step. I'm going to defer the more technical statement for UT to Step2, but for now we could state in this meta statement as: UT <-> "It's impossible to know the exact inductive-distribution of infinitely many numbers with non-inductive primeval properties". Again, we have to be clear by what we mean by "exact", "inductive-distribution", and "non-inductive", "primeval" and we'll define them in next step. For now, by inductive-distribution we mean the distribution formula: Axyz[x*(y+z)=(x*y) + (x*z)] would have to have is shortest proof as a _non-trivial and by-induction_ proof.
From: Nam Nguyen on 13 Feb 2010 11:44 Nam Nguyen wrote: > First let's assume the following languages: > > - L1 = L1(0,S,+) > - L2 = L2(0,<,*) > - L = L(0,S,+,*,<) > > Now the steps: > > Step2: Demonstrate that since L is the combined language of L1 and L2, the > undefinability of the exactness mentioned in Step 1 will transform > into a an arithmetic truth value that if being "true" would be > impossible to know. Also, in Step1 we deferred detailed explanations of the following meta UT statement to this step 2: > UT <-> "It's impossible to know the exact inductive-distribution of > infinitely many numbers with non-inductive primeval properties". Let's now present 2 technical versions of the above UT: UT1 <-> It's impossible to know the (arithmetic) truth value of a formula of the following form, if it (the formula) is true: (AI)primeval1(p,x) UT2 <-> It's impossible to know the (arithmetic) truth value of a formula of the following form, if it (the formula) is true: (AI)primeval2(c,x) where: primeval1(p,x) <-> (prime(p) /\ Ey[x=p*y] /\ (x >= p*p)) -> Ex1...xn[AND(x:1-n)prime(x) /\ (x=iFunc1(x1,...,xn))] where: p is a constant, n=p, and AND(x:1-n)P(x) <-> P(x1) /\ P(x2) /\ ... P(xn) iFunc1(x1,...,xn) <-> a function that would involve only 'S' or +' as non-logical symbols. primeval2(c,x) <-> prime(x) -> (c=iFunc2(x)), where c is a constant. iFunc2(x) <-> a function that must involve 'S' or '+' as non-logical symbols. prime(x) <-> previously defined in L2 (AI)P <-> previously defined in L2, for a property P(x). An example of UT1 is the familiar Goldbach's Conjecture: "There are infinitely many examples Goldbach's Conjecture." An example of UT2 would be: "There are infinitely many primes each of which is 1, modulo the sum of all its digits, except the digit next to the last one."
From: Nam Nguyen on 13 Feb 2010 11:51 Nam Nguyen wrote: > Nam Nguyen wrote: > >> First let's assume the following languages: >> >> - L1 = L1(0,S,+) >> - L2 = L2(0,<,*) >> - L = L(0,S,+,*,<) >> >> Now the steps: >> >> Step2: Demonstrate that since L is the combined language of L1 and L2, >> the >> undefinability of the exactness mentioned in Step 1 will transform >> into a an arithmetic truth value that if being "true" would be >> impossible to know. > > Also, in Step1 we deferred detailed explanations of the following meta UT > statement to this step 2: > >> UT <-> "It's impossible to know the exact inductive-distribution of >> infinitely many numbers with non-inductive primeval properties". > > Let's now present 2 technical versions of the above UT: > > UT1 <-> It's impossible to know the (arithmetic) truth value of a formula > of the following form, if it (the formula) is true: > > (AI)primeval1(p,x) > > > UT2 <-> It's impossible to know the (arithmetic) truth value of a formula > of the following form, if it (the formula) is true: > > (AI)primeval2(c,x) > > where: > > primeval1(p,x) <-> (prime(p) /\ Ey[x=p*y] /\ (x >= p*p)) -> > Ex1...xn[AND(x:1-n)prime(x) /\ (x=iFunc1(x1,...,xn))] > > where: p is a constant, n=p, and > AND(x:1-n)P(x) <-> P(x1) /\ P(x2) /\ ... P(xn) > > iFunc1(x1,...,xn) <-> a function that would involve only 'S' or +' as > non-logical symbols. > primeval2(c,x) <-> prime(x) -> (c=iFunc2(x)), where c is a constant. > iFunc2(x) <-> a function that must involve 'S' or '+' as > non-logical > symbols. > prime(x) <-> previously defined in L2 > (AI)P <-> previously defined in L2, for a property P(x). > > An example of UT1 is the familiar Goldbach's Conjecture: Sorry. I meant "An example of UT1 is related to the familiar Goldbach's Conjecture:" > > "There are infinitely many examples Goldbach's Conjecture." > > An example of UT2 would be: > > "There are infinitely many primes each of which is 1, modulo the sum > of all its digits, except the digit next to the last one." >
From: Nam Nguyen on 13 Feb 2010 12:37
Nam Nguyen wrote: > Nam Nguyen wrote: > >> First let's assume the following languages: >> >> - L1 = L1(0,S,+) >> - L2 = L2(0,<,*) >> - L = L(0,S,+,*,<) >> >> Now the steps: >> >> Step2: Demonstrate that since L is the combined language of L1 and L2, >> the >> undefinability of the exactness mentioned in Step 1 will transform >> into a an arithmetic truth value that if being "true" would be >> impossible to know. > > Also, in Step1 we deferred detailed explanations of the following meta UT > statement to this step 2: > >> UT <-> "It's impossible to know the exact inductive-distribution of >> infinitely many numbers with non-inductive primeval properties". > > Let's now present 2 technical versions of the above UT: > > UT1 <-> It's impossible to know the (arithmetic) truth value of a formula > of the following form, if it (the formula) is true: > > (AI)primeval1(p,x) > > > UT2 <-> It's impossible to know the (arithmetic) truth value of a formula > of the following form, if it (the formula) is true: > > (AI)primeval2(c,x) > [I'd like to re-edit the following for a correction on 'primeval2(c,x)' and on UT2 example.] where: primeval1(p,x) <-> (prime(p) /\ Ey[x=p*y] /\ (x >= p*p)) -> Ex1...xn[AND(x:1-n)prime(x) /\ (x=iFunc1(x1,...,xn))] where: p is a constant, n=p, and AND(x:1-n)P(x) <-> P(x1) /\ P(x2) /\ ... P(xn) iFunc1(x1,...,xn) <-> a function that would involve only 'S' or +' as non-logical symbols. primeval2(c,x) <-> (prime(x) /\ AND(x:1-n)P(x)) -> (c=iFunc2(x)), where c is a constant and, here, AND(x:1-n)P(x) <-> P1(x) /\ ... Pn(x) for some arbitrary P1, ..., Pn(x). iFunc2(x) <-> a function that must involve 'S' or '+' as non-logical symbols. prime(x) <-> previously defined in L2. (AI)P <-> previously defined in L2, for a property P(x). An example of UT1 is: "There are infinitely many examples GC." An example of UT2 is: "There are infinitely many primes each corresponds to a counter- example of GC and each is 1, modulo the sum of all its digits, except the digit next to the last one." |