Peano arithmetic is enough, because Peano arithmetic encodes computation
$\begingroup$ PA is enough, because PA can encode computation. This is longer than I expected, and was made longer still by some browser crashes. But I'd been idly thinking of writing these ideas up. I hadn't for these reasons. It is a lot of work. What I have to say is obvious to logicians, and they would consider the detour into programming to only be a distraction. Computer programmers who can appreciate the programming detour, are mostly not that interested in logic. But this question hi