Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-3/12j/e/solution

If were total recursive, its recursion-theorem fixed point would satisfy when and when , a contradiction in either case. The same argument applies to . A total recursive reduction with would provide such a forbidden switch (and would collapse the known distinct arithmetical complexities), so none exists.

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