All Questions
1 question
12
votes
1
answer
835
views
Transfinitely extending $\sf PA$ — can we get stronger than $\sf ZFC$?
Let $\sf PA$ denote the theory of natural numbers with constants $(0, 1)$ and binary operators $(+,\times)$ based on the first-order predicate calculus with equality, having the following axioms, ...
- 6,829