4 ms·
ZF is subject to the incompleteness theorems, it can't prove Con(ZF). Even much simpler systems, like Peano arithmetic, are subject to them.
by cousin_it 10y ago
ZF is subject to the incompleteness theorems, it can't prove Con(ZF). Even much simpler systems, like Peano arithmetic, are subject to them.
- Mathnerd314 10y agoIt's consistent/complete in first-order logic, because Con(ZF) is a higher-order statement and thus can't even be formulated. First-order logic has a lot of nice properties, e.g. "if a result is finitary in the sense that it can be phrased as a first-order statement in Peano Arithmetic, and it can be proven using the axiom of choice (or more precisely in ZFC set theory), then it can also be proven without the axiom of choice (i.e. in ZF set theory)." (c.f. https://terrytao.wordpress.com/2013/12/07/ultraproducts-as-a-bridge-between-discrete-and-continuous-analysis/ https://terrytao.wordpress.com/2013/12/07/ultraproducts-as-a...)
- cousin_it 10y agoCon(ZF) is a sentence in ZF.