(before saying it's duplicate, read whole question)

I was told by someone that we can define addition and multiplication purely in terms of successor function, provided that we work in second order arithmetic. When trying to find a way in which it's defined, I found this M.SE question, and in the first answer a definition is given. I found it, however, quite unsatisfactory, because this definition uses quantification over functional symbols (which can be rewritten to use binary predicate quantification), and I expected a definition which would only require quantification over subsets of $\Bbb N$ (equivalently unary predicates). I expect something similar to answer to this question.

My question is: how to define addition in second order arithmetic using quantification over sets only? Is that even possible?

Thanks in advance.


As it says in the answer to Can equinumerosity by defined in monadic second-order logic?, Büchi showed that the monadic second order of theory of the successor function is decidable. This implies that you cannot define addition in this theory, because if you could, then you could also define multiplication (as shown in the answer to How to define multiplication in addition terms in monadic second order logic?) and then the theory would be at least as strong as PA, which is not decidable.

I can't find the original work by Büchi online, but this paper by Rabinovich and Thomas has the relevant references.

  • $\begingroup$ I must admit that I'm a bit disappointed with the impossibility of this result. Thanks for the reference. $\endgroup$ – Wojowu Jan 3 '15 at 13:05

Not sure what you mean by "using quantification over sets only," but this is how you define addition and multiplication in terms of the successor function.


$\forall x \in N: x+0=x$

$\forall x,y\in N: x+S(y) = S(x+y)$


$\forall x \in N: x\cdot0=0$

$\forall x,y\in N: x\cdot S(y) = x\cdot y + x$

You can construct these functions using Peano's axioms and the axioms of set theory.

For the addition function, we would define the set of ordered triples $A$ such that

$\forall x,y,z:[(x,y,z)\in A \iff (x,y,z)\in N^3 $

$\land \forall B\subset N^3:[\forall a\in N: (a,0,a)\in B \land \forall a,b,c: [ (a,b,c)\in B \implies (a,S(b),S(c))\in B]$

$ \implies (x,y,z)\in B]] $

Similarly, for the multplication function (having defined +), we would define the set of ordered triples $M$ such that

$\forall x,y,z:[(x,y,z)\in M \iff (x,y,z)\in N^3 $

$\land \forall B\subset N^3:[\forall a\in N: (a,0,0)\in B \land \forall a,b,c: [ (a,b,c)\in B \implies (a,b+1,c+a\in B]$

$ \implies (x,y,z)\in B]] $

  • $\begingroup$ Definition in second order logic would be a formula P(x,y,z), which importantly doesn't involve itself, which is true iff x+y=z. $\endgroup$ – Wojowu Jan 14 '15 at 12:42
  • $\begingroup$ @Wojowu See edit (below line). $\endgroup$ – Dan Christensen Jan 14 '15 at 17:00
  • $\begingroup$ Language of second order arithmetic doesn't allow us to work with ordered triples. $\endgroup$ – Wojowu Jan 14 '15 at 17:03
  • $\begingroup$ @Wojowu If so, it would indeed appear to be impossible. $\endgroup$ – Dan Christensen Jan 14 '15 at 17:13
  • $\begingroup$ @Wojowu: in non-monadic second-order logic, we could quantify over ternary relations (or, more generally, over any type of finitary relation). But the question you asked is clearly about monadic logic, where we can only quantify over unary relations. $\endgroup$ – Carl Mummert Jan 15 '15 at 11:50

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