Absoluteness Of Consistency

Note that any arithmetic operation is an Absolute Operation. Take from Gödel’s Incompleteness Theorems. Note that is an arithmetic function. Thus is absolute for Transitive Models. Let and define

Also let

Proposition

for all

Proof

If there is a model of , then there is certainly no proof of Thus:

Let be a Transitive Model of . is Absolute so

i.e.

But then we have proved

(because it has a model, namely ) Now is a Transitive Model of . We continue by induction.

Remark

We have proved that is much stronger than . However, assuming , take a Model of We can construct an inner model (e.g. Constructible Model of Set Theory)

where is Transitive in . This might seem like an issue, as we might think that . However, doesn’t see as a set. Thus doesn’t entail (at least not as witnessed by ) and certainly doesn’t entail .