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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.