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.

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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.