User blog comment:LittlePeng9/First order oodle theory/@comment-35470197-20180827041403

> We also assume that oodles satisfy all the properties which are considered "natural" properties, e.g. extensionality or existence of full power set.

Could you precisely tell me whole axioms? I think that \(\textrm{FOOT}\) is not well-defined under a "natural" axiom. For example, how do you formalise the definition of \(\alpha_0\)?

If you formalise the definition of \(\alpha_0\), say a formula \(\varphi(\alpha)\), then the existence of \(\alpha_0\) satisfying \(\varphi(\alpha_0)\) ensures that the existence of \(\beta < \alpha_0\) satisfying \(\varphi(\beta)\). By the minimality of \(\alpha_0\), you obtain \(\alpha_0 = \beta < \alpha_0\).

> Now we have to verify that formula \(\varphi(A)\), for \(A \in V_{\textrm{Ord}}\) is true in \(V\) iff it is in \(V_{\textrm{Ord}}\).

So for any closed formula \(\varphi\), \(V_{\textrm{Ord}} \models \varphi\) is equivalent to \(\varphi\). So let \(\varphi\) denote the existence of \((\alpha_n)_{n < \omega}\), which is true under the "natural" axiom which you assumed on \(V\). Then \(V_{\textrm{Ord}}\) satisfies \(\varphi\), and hence \((\alpha_n^{V_{\textrm{Ord}}})_{n < \omega}\) is well-defined.

On the other hand, \(\alpha_0^{V_{\textrm{Ord}}}\) satisfies the same property as \(\alpha_0\). Indeed, for any parameter-free formula \(\phi\), \(\phi^{V_{\textrm{Ord}}}\) is equivalent to \(\phi\) by your argument. Many formulae, e.g. \(\beta \in \alpha\), "\(\alpha\) is oodle", "\(n\) is the Goedel number of a formula", and so on, is absolute with respect to the inclusion \(V_{\textrm{Ord}} \hootorightarrow V\) again by your argument. So, it implies \(\alpha_n^{V_{\textrm{Ord}}} = \alpha_n\) for any \(n < \omeg\). You obtain \(\textrm{Ord}^{V_{\textrm{Ord}}} = \textrm{Ord}\). It is an obvious contradiction.