4

Presburger Arithmetic is decidable theory but weaker than Peano Arithmetic. Are there systems in some sense that are:

  1. stronger than Presburger but weaker than Peano and remain decidable?

  2. weaker than Peano but stronger than Presburger and remain undecidable but all their undecidable statements are not decidable to be undecidable?

  3. stronger than Presburger but weaker than Peano and remain undecidable but all their undecidable statements are decidable to be undecidable?

Turbo
  • 6,319
  • 2
    In points 2 and 3, what does it mean for an undecidable statement to be "decidable to be undecidable"? – user473884 May 15 '18 at 17:14
  • are you really donald trump? – Turbo May 15 '18 at 17:15
  • 1
    It doesn't make sense to say that a statement is "not decidable to be undecidable". – Rob Arthan May 15 '18 at 21:18
  • In addition to the issues with "decidability of undecidability", you should clarify what exactly you mean by "stronger". There are multiple ways of comparing the strength of theories in different languages. – Alex Kruckman May 16 '18 at 13:33

1 Answers1

0

I asked a similar question and got some great answers.

One example is that we can add a relation $\vert_2$ to Presburger arithmetic with the interpretation $x \vert_2 y \leftrightarrow \exists_{z,w}(2^z=x \land x \cdot w = y)$ and the theory stays decidable.

Dan Brumleve
  • 18,041
  • Interesting. Does that mean in Presburger arithmetic statements like $2x_1+5x_4=2^y$ will be allowed? – Turbo May 17 '18 at 15:56
  • 1
    No, the idea is that we can only use the symbol $\vert_2$ so we can't even say "$x$ is $2$ to the power of $y$", only things like "$x$ is a power of $2$" and "$x$ and $y$ are both powers of $2$ and $y$ is bigger". And we need some axioms to relate $\vert_2$ to $+$ but I'm not sure what those are exactly (I'm not sure if there is any complete finite axiomatization but otherwise I suppose we can add all the true statements and have a decidable theory). – Dan Brumleve May 17 '18 at 16:13