Lemma 2.2.3. For any natural number \(n\) and \(m\), \(n+(m\pp)=(n+m)\pp\).
\(Proof~by~Tao\). We induct on \(n\) (keeping \(m\) fixed). We first consider the base case \(n=0\). In this case we have to prove \(0+(m\pp)=(0+m)\pp\). But by definition of addition, \(0+(m\pp)=m\pp\) and \(0+m=m\), so both sides are equal to \(m\pp\) and are thus equal to each other. Now we assume inductively that \(n+(m\pp)=(n+m)\pp\); we now have to show that \((n\pp)+m(\pp)=((n\pp)+m)\pp\). The left-hand side is \((n+(m\pp))\pp\) by definition of addition, which is equal to \(((n+m)\pp)\pp\) by the inductive hypothesis. Similarly, we have \((n\pp)+m=(n+m)\pp\) by the definition of addition, and so the right-hand side is also equal to \(((n+m)\pp)\pp\). Thus both sides are equal to each other, and we have closed the induction. \(\square\)