I have an abstract syntax for Presburger arithmetic, along with a fixpoint function determining a given formula's propositional denotation (you can see it here: https://gist.github.com/d4hines/d9a0c674f324cab46d2cf0967bde1ac3).
I'd like to prove that the truth value of any given formula is decidable. Since it's Presburger arithmetic, I know it must decidable. I've heard that the decision procedures for Presburger arithmetic are very complicated. I'd like to reuse the existing one in Coq.
How can I do this?
Thanks!
There are a few reasons why lia will not be of great help to you.
A small true goal like exists x : Z, 2 < x < 4 is not solved by lia: this tactic is not complete for Prestburger arithmetic
Even if lia was complete for Presburger, it would act as an oracle: giving you an answer every time you need one for a true formula. But when presented with a false formula, lia only says to Coq I can't do it, it does not say I have a proof that this can't be done. In other words, the information that a proof procedure is complete may not be stored in the Coq system as a Coq readable proof.