As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn can be decided). As a consequence, Presburger arithmetic is consistent, because it can easily be computed that 0 = 1 is not a theorem.
I am wondering how much first-order arithmetic is required to prove Presburger arithmetic admits quantifier elimination. We know Robinson's Q is not strong enough to prove Con(Presburger); how much more is required?
