Comment by contraposit

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)

Even before I started the video, I had a feeling it was going to lead to a kind of "introspective" mathematics that can reason about its own reasoning. I was not disappointed, thank you.

Physics, Topology, Logic and Computation: A Rosetta Stone - https://arxiv.org/abs/0903.0340