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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 (r
by gottheUIblues 6mo ago
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)
- contraposit 6mo agoI am referring to this slide : https://youtu.be/EVwQsvof7Hw?t=1646 https://youtu.be/EVwQsvof7Hw?t=1646