 Solution

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

Fix

m

and induct on

n

. The initial equality

m + 0 = 0 + m

follows from the defining zero clause and (ii). Assuming

m + n = n + m

, the successor clause and (iii) give

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

This proves

m + n = n + m (m, n \in ω) .

The extension to arbitrary ordinals is false. By the limit clause for ordinal addition,

1 + ω = sup_{n < ω} (1 + n) = ω,

whereas

ω + 1

ω

and is strictly greater than it. Therefore

1 + ω \neq = ω + 1.

Finite addition is commutative; ordinal addition records the order of concatenated well-orders and need not be commutative.

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