alt Hacker NewsErr? Peano Arithmetic is provably consistent in ZFC, but it is not in itself (if PA is consistent). Therefore if PA is consistent it is not equivalent to ZFC (regardless of whether ZFC is consistent or not)
Err? Peano Arithmetic is provably consistent in ZFC, but it is not in itself (if PA is consistent). Therefore if PA is consistent it is not equivalent to ZFC (regardless of whether ZFC is consistent or not)
I am referring to this slide : https://youtu.be/EVwQsvof7Hw?t=1646