A Finite Model of Peano Arithmetic
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From: RussellE on 3 Mar 2010 22:27 On Mar 3, 7:19 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Mar 3, 6:10 pm, RussellE <reaste...(a)gmail.com> wrote: > > > Start with Peano's axioms:http://en.wikipedia.org/wiki/Peano_axioms. > > If an axiomatic theory is consistent, it will still be consistent if > > we > > remove an axiom. > > > I will remove axiom 7: 0 is not the successor of any natural number. > > I can now define the set {0,1} and show it is a model of PA. > > > Define a successor function: > > S(0) = 1 > > S(1) = 0 > > As near as I can tell, this set is a model of PA-Axiom 7. > > So then, this is not a model for Peano Arithmetic, is it? This is a model of Peano arithmetic. It satisfies the 15 axioms of Peano arithmetic given in the Wiki article. > Hell, if > you want "Peano - "0 is not the successor anyone"", then you can take > {0} and S:{0}-->{0} be the identity. So what? This isn't a model for Peano arithmetic because it doesn't have an identity for multiplication. Russell - Integers are an illusion
From: RussellE on 4 Mar 2010 01:41 On Mar 3, 4:10 pm, RussellE <reaste...(a)gmail.com> wrote: > Start with Peano's axioms:http://en.wikipedia.org/wiki/Peano_axioms. > If an axiomatic theory is consistent, it will still be consistent if > we > remove an axiom. > > I will remove axiom 7: 0 is not the successor of any natural number. > I can now define the set {0,1} and show it is a model of PA. > > Define a successor function: > S(0) = 1 > S(1) = 0 > > As near as I can tell, this set is a model of PA-Axiom 7. I can define a set of natural numbers and a successor function that is consistent with every axiom of PA except one. I define 0 to be the successor of a natural number. My theory even has induction. I can define a model of Peano arithmetic with N, 0, and successor. I can show this model is consistent with the axioms of Peano arithmetic. I have often seen Godel's Incompleteness theorem describe as no formal theory capable of defining Peano arithmetic can be both consistent and complete. I am fairly sure my theory can be shown to be equivalent to FOL. Godel proved FOL is both consistent and complete in 1929. http://en.wikipedia.org/wiki/G%C3%B6del's_completeness_theorem I have also read that FOL is incapable of defining arithmetic. Assume my theory is equivalent to FOL. Then there is a method to formulate arithmetic in FOL and FOL is inconsistent by the Incompleteness theorem. If my theory is not equivalent to FOL, then some undecidable statement must be derivable in my theory. How can there be undecidable statements about arithmetic in a finite system? Russell - the universe is one dimensional
From: Aatu Koskensilta on 4 Mar 2010 01:44 RussellE <reasterly(a)gmail.com> writes: > I have Peano arithmetic and I can prove the system is > consistent. Doesn't this prove PA is inconsistent? No. What argument do you have in mind? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 4 Mar 2010 01:52 RussellE <reasterly(a)gmail.com> writes: > I have often seen Godel's Incompleteness theorem describe as no formal > theory capable of defining Peano arithmetic can be both consistent and > complete. You may well have seen such a description. Setting the obvious incorrectness of this description to one side, we may observe that in it "Peano arithmetic" most likely refers to Peano arithmetic, not Peano arithmetic with an axiom removed. > I am fairly sure my theory can be shown to be equivalent to FOL. Godel > proved FOL is both consistent and complete in 1929. > http://en.wikipedia.org/wiki/G%C3%B6del's_completeness_theorem Alas, you are the victim of an elementary misunderstanding: the consistency of first-order logic is a triviality; first-order logic is not complete in the sense of the incompleteness theorems. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: RussellE on 4 Mar 2010 02:09
On Mar 3, 10:52 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > RussellE <reaste...(a)gmail.com> writes: > > I have often seen Godel's Incompleteness theorem describe as no formal > > theory capable of defining Peano arithmetic can be both consistent and > > complete. > > You may well have seen such a description. Setting the obvious > incorrectness of this description to one side, we may observe that in it > "Peano arithmetic" most likely refers to Peano arithmetic, not Peano > arithmetic with an axiom removed. If Peano arithmetic is consistent, it is also consistent when we remove one axiom. > > I am fairly sure my theory can be shown to be equivalent to FOL. Godel > > proved FOL is both consistent and complete in 1929. > >http://en.wikipedia.org/wiki/G%C3%B6del's_completeness_theorem > > Alas, you are the victim of an elementary misunderstanding: the > consistency of first-order logic is a triviality; Then, why was it a big deal when Godel proved FOL is consistent and complete? > first-order logic is > not complete in the sense of the incompleteness theorems. Can you give an example of an undecidable statement in FOL? Can you express "this statement can't be proven" in FOL? I still find it hard to believe there are undecidable statements about arithmetic for a finite set of natural numbers. Russell - 2 many 2 count |