When I was an undergraduate I had a philosophy professor tell me that if you took the axioms of number theory with only addition or only multiplication then the resulting theory was decidable. It is only after we have both a multiplication and addition that incompleteness results. I looked briefly online, but found no mention of this. Is it true? Is there a reference or easy proof of this fact?
2026-03-27 21:19:16.1774646356
Completeness of a Sub Theory of Number Theory
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Yes. As @coffeemath notes, Presburger arithmetic (think: first-order Peano Arithmetic with the language restricted to successor and addition) is decidable and complete. This was proved before Gödel showed, in effect, that Peano Arithmetic is incomplete (one more reason that the first incompleteness theorem was a surprise! -- after all, we might ask, why should including multiplication as well make all the difference?).
There's a nice proof that Presburger arithmetic is complete (and hence decidable) in Ch. 7 of Alec Fisher's Formal number theory and computability (Vol. 7 of the Oxford Logic Guides). It uses a standard technique called "elimination of quantifiers".