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Aug 23 at 16:01 answer added L. Gitin timeline score: 3
Sep 3, 2015 at 14:35 comment added Mary Star I am looking for the proof that the existential theory of $\mathbb{Z}$ with the structure of addition, divisibility and the relation $(\exists s \in \mathbb{Z})m=np^s$ is undecidable. I have found one proof at the paper of Denef and I was wondering if I can find the proof also somewhere else... Do you maybe where I can find it? @DávidTóth
Sep 3, 2015 at 12:37 comment added Dávid Natingga I think I do not understand your question. The statement $\exists s \in \mathbb{Z} (m=np^s)$ is decidable easily: just check all $s$ up to the values of $m$.
Sep 2, 2015 at 21:43 comment added Gerhard Paseman Based on mathoverflow.net/questions/216251/… , my guess is it is a binary relation (as p is fixed), and that this question will not have a concise and suitable answer. Gerhard "Unsure And Undecided About Suitability" Paseman, 2015.09.02
Sep 2, 2015 at 21:08 comment added Mary Star I found the relation $(\exists s \in \mathbb{Z})m=np^s$ in the paper jstor.org/stable/2275396?seq=1#page_scan_tab_contents . I added how it is formulated in the paper of Denef. @JoelDavidHamkins
Sep 2, 2015 at 21:04 history edited Mary Star CC BY-SA 3.0
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Sep 2, 2015 at 20:52 comment added Joel David Hamkins Could you clarify the relation? Do you mean a binary relation, where $p$ is a fixed prime? Or do you mean a trinary relation, or what?
Sep 2, 2015 at 20:37 history asked Mary Star CC BY-SA 3.0