 Solution

ID: past-exam-of-the-logic-and-set-theory-course-of-the-university-of-oxford/2007/b/5/iii/solution

Fix

m

and induct on

n

. At

n = 0

m^{+} + 0 = m^{+} = (m + 0)^{+} = m + 0^{+} .

m^{+} + n = m + n^{+}

, then

m^{+} + n^{+} = (m^{+} + n)^{+} = (m + n^{+})^{+} = m + (n^{+})^{+} .

The last expression is the right-hand side required at the successor

n^{+}

. Therefore

m^{+} + n = m + n^{+} (m, n \in ω) .

In particular both sides also equal

(m + n)^{+}

by the recursion equation for the second argument.

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