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Location: Chapter 24, solutions to Problems 6 and 7

Problem 6

It is (formula given in the solution)

... such that for every n and m, A\overline{n}\,\overline{m}=(Z\,\overline{m})\overline{0}((\oplus)(A(\overline{n}(P\,\overline{m}))\overline{n})). Again, such a bird A can be ...

It should become

... such that for every n and m, A\overline{n}\,\overline{m}=Z\,\overline{m}\overline{0}(\oplus(A\overline{n}(P\,\overline{m}))\overline{n}). Again, such a bird A can be ...

Problem 7

It is (two expressions to correct)

Note: the exponentiating bird is represented in the following with the symbol \varepsilon.

... for any positive number m, \varepsilon\,\overline{n}\,\overline{m}=\otimes(\varepsilon\,\overline{n}\,\overline{m})\overline{n}. Equivalently we want a bird \varepsilon such that for all n and m, \varepsilon\,\overline{n}\,\overline{m}=Z\,\overline{m}\overline{1}(\otimes(\varepsilon\,\overline{n}\,\overline{m})\overline{n}). Again, such a bird \varepsilon can be ...

It should become

... for any positive number m, \varepsilon\,\overline{n}\,\overline{m}=\otimes(\varepsilon\,\overline{n}(P\,\overline{m}))\overline{n}. Equivalently we want a bird \varepsilon such that for all n and m, \varepsilon\,\overline{n}\,\overline{m}=Z\,\overline{m}\overline{1}(\otimes(\varepsilon\,\overline{n}(P\,\overline{m}))\overline{n}). Again, such a bird \varepsilon can be ...

Short explanation

In the correction to Problem 6, besides a few redundant parentheses that can go away just for ease of reading, the crucial point is the term A\overline{n}(P\,\overline{m}): with the formula given in the text, A would be applied, incorrectly, to the two terms (\overline{n}(P\,\overline{m})) and \overline{n}, leaving the sum bird to act on a single term. The correction restores the intended interplay between \oplus and A.

As for Problem 7, the expressions simply forgot to decrease m by one when "unfolding" one step in the recursive definition of the exponentiation.