PA subsets consistency

Question

IF i have T in piano arithmetic and suppose I add a new symbol c

suppose it is T1

T1=T U {c>1,c>1+1,c>1+1+1.......}

Is it true that every finite subset of T1 is consistent and can we show that a structure M |= T1

I have read in a notes that it is by soundness we can confirm that it is consistent

From my googling I understand that if aal the subsets are consistent then T1 is consistent , how can I show that all the subsets of T1 are consistent and the existence of the structure

score 1 · Answer 1 · answered Sep 22 '15 at 02:46

1

For any finite subset of the axioms there is one with the most $1$s on the right. Any model of PA will then satisfy PA plus a finite number of these axioms by taking $c$ larger than this number. This shows that every finite subset of $T1$ has a model (assuming PA has a model) and is therefore consistent. The compactness theorem then promises that the whole set of axioms is consistent and has a model, but the usual model of PA doesn't work, so there must be a model that satisfies PA but is not the usual model.

answered Sep 22 '15 at 02:46

Ross Millikan

374,822

Good use of the Compactness Theorem – DanielWainfleet Sep 22 '15 at 03:51
then suppose we Consider the closed terms that do not use the constant c (such as (1 +1)
(1+1+1) ). Do these terms represent standard or non-standard elements of M
– Maaya Sep 22 '15 at 11:02
They will all represent standard elements, as you can represent them in the standard model. – Ross Millikan Sep 22 '15 at 13:44

PA subsets consistency

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