Past exam of the logic and set theory course of the University of Oxford / 2007 / B / 5 / iv

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- Solution 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.

For the induction step, use the multiplication recursion, the induction hypothesis, and addition associativity from (i):

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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