 Solution

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

Fix

m, n

and induct on

k

. For

k = 0

m \cdot (n + 0) = m \cdot n = m \cdot n + 0 = m \cdot n + m \cdot 0.

m \cdot (n + k^{+}) = m \cdot (n + k)^{+} = m \cdot (n + k) + m = (m \cdot n + m \cdot k) + m = m \cdot n + (m \cdot k + m) = m \cdot n + m \cdot k^{+} .

Thus

m \cdot (n + k) = m \cdot n + m \cdot k .

This establishes the indicated distributive law without assuming multiplication is commutative.

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