2. Let T be a theory in a recursively numbered language, and assume that there is an interpretation of \text{Cn }A_E into T. Show that T is strongly undecidable; i.e., whenever T^\prime is a theory in the language for which T \cup T^\prime is consistent, then \sharp T^\prime is not recursive.


This is my first attempt to the sketch of the proof based on Theorem 37D. Could you check the following?

Let \pi be such an interpretation of \text{Cn }A_E into \Pi for which \Pi \subseteq T. Let \Delta be the consistent theory \text{Cn}(T^\prime \cup \Pi). Let \Delta_0 be the corresponding theory \pi^{-1}[\Delta] in the language of number theory. Since \Delta is the consistent theory, \Delta_0 is a consistent theory such that A_E \subseteq \Delta_0 (from Section 2.7). By Theorem 35C, \sharp \Delta_0 is not recursive, which follows that \sharp \Delta is not recursive (by the arguments in page 273).

We also have \tau \in \Delta iff (\phi \rightarrow \tau) \in T^\prime, where \phi is the conjunction of the members of \Pi.

Therefore, T^\prime is not recursive, nor is \sharp T^\prime.