A Finite Model of Peano Arithmetic
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From: RussellE on 3 Mar 2010 19:10 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. Russell - 2 many 2 count
From: Aatu Koskensilta on 3 Mar 2010 19:16 RussellE <reasterly(a)gmail.com> writes: > As near as I can tell, this set is a model of PA-Axiom 7. Great. An obvious question suggests itself: so what? -- 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 3 Mar 2010 20:28 On Mar 3, 4:16 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > RussellE <reaste...(a)gmail.com> writes: > > As near as I can tell, this set is a model of PA-Axiom 7. > > Great. An obvious question suggests itself: so what? I have Peano arithmetic and I can prove the system is consistent. Doesn't this prove PA is inconsistent? Russell - 2 manay 2 count
From: Arturo Magidin on 3 Mar 2010 22:19 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? 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? -- Arturo Magidin
From: Arturo Magidin on 3 Mar 2010 22:20
On Mar 3, 7:28 pm, RussellE <reaste...(a)gmail.com> wrote: > On Mar 3, 4:16 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > > RussellE <reaste...(a)gmail.com> writes: > > > As near as I can tell, this set is a model of PA-Axiom 7. > > > Great. An obvious question suggests itself: so what? > > I have Peano arithmetic No, you don't, because your set does *NOT* satisfy he Peano axioms, so it is not "Peano arithmetic". >and I can prove the system > is consistent. Doesn't this prove PA is inconsistent? Since you do *not* have Peano Arithmetic, yout implication has a false antecedent and therefore is true. > - 2 many 2 count Your errors and misconceptions, or just your wild claims? -- Arturo Magidin |