{\displaystyle {\begin{aligned}a+1&=a+S(0)&{\mbox{by definition}}\\&=S(a+0)&{\mbox{using (2)}}\\&=S(a),&{\mbox{using (1)}}\\\\a+2&=a+S(1)&{\mbox{by definition}}\\&=S(a+1)&{\mbox{using (2)}}\\&=S(S(a))&{\mbox{using }}a+1=S(a)\\\\a+3&=a+S(2)&{\mbox{by definition}}\\&=S(a+2)&{\mbox{using (2)}}\\&=S(S(S(a)))&{\mbox{using }}a+2=S(S(a))\\{\text{etc.}}&\\\end{aligned}}}