Proposition 2.2.4 (Addition is commutative). For any natural numbers \(n\) and \(m\), \(n+m=m+n\).
\(Proof~by~Tao\). We shall use induction on \(n\) (keeping \(m\) fixed). First we do the base case \(n=0\), i.e., we show \(0+m=m+0\). By the definition of addition, \(0+m=m\), while by Lemma 2.2.2, \(m+0=m\). Thus the base case is done. Now suppose inductively that \(n+m=m+n\), now we have to prove that \((n\pp)+m=m+(n\pp)\) to close the induction. By the definition of addition, \((n\pp)+m=(n+m)\pp\). By Lemma 2.2.3, \(m+(n\pp)=(m+n)\pp\), but this is equal to \((n+m)\pp\) by the induction hypothesis \(n+m=m+n\). Thus \((n\pp)+m=m+(n\pp)\) and we have closed the induction. \(\square\)