Lemma: Suppose that φ is a Regular Φ axiom system that is consistent. Then φ must be unable to prove the sentence Ω(φ).

Can someone explain to me why Ω(φ) must be a true sentence?

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- Apr 27th 2010, 02:23 PMjjbrianIncompleteness thereom
Lemma: Suppose that φ is a Regular Φ axiom system that is consistent. Then φ must be unable to prove the sentence Ω(φ).

Can someone explain to me why Ω(φ) must be a true sentence? - Apr 28th 2010, 10:15 AMemakarov
There are a couple of non-universal notations here. What is a "Regular Φ axiom system", namely what does Φ mean? And $\displaystyle \Omega(\phi)$ is the formula stating consistency of $\displaystyle \phi$, something that is also denoted $\displaystyle \text{Con}(\phi)$ or $\displaystyle \text{Consis}(\phi)$?

Presumably, $\displaystyle \Omega(\phi)$ says something "there is no $\displaystyle \phi$-derivation of 0=1". If it is false (in the standard model $\displaystyle \mathbb{N}$), then there exists, in fact, a $\displaystyle \phi$-derivation of 0=1. Then $\displaystyle \phi$ is contradictory and therefore derives everything. However, it is proved that $\displaystyle \phi$ does not prove $\displaystyle \Omega(\phi)$.