Prove the identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ for all natural numbers $a,b$.
The following derivation aims to justify every step and avoid seemingly obvious jumps.
$$ \begin{alignat*}{3} (a+b)^{2} & =(a+b)^{1++} & \hspace{4cm} & & {1++=2}\\ \\ & =(a+b)^{1}\times(a+b) & \hspace{4cm} & & \text{Definition 2.3.11}\\ \\ & =(a+b)^{0++}\times(a+b) & \hspace{4cm} & & {0++=1}\\ \\ & =(a+b)^{0}\times(a+b)\times(a+b) & \hspace{4cm} & & \text{Definition 2.3.11}\\ \\ & =1\times(a+b)\times(a+b) & \hspace{4cm} & & \text{Definition 2.3.11}\\ \\ & =(0++)\times(a+b)\times(a+b) & \hspace{4cm} & & {0++=1}\\ \\ & =((0\times(a+b)\times(a+b))+(a+b)\times(a+b) & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =(a+b)\times(a+b) & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =a\times(a+b)+b\times(a+b) & \hspace{4cm} & & \text{Proposition 2.3.4}\\ \\ & =a\times a+a\times b+b\times a+b\times b & \hspace{4cm} & & \text{Proposition 2.3.4}\\ \\ & =a\times a+a\times b+a\times b+b\times b & \hspace{4cm} & & \text{Lemma 2.3.2}\\ \\ & =a^{2}+(a\times b)+(a\times b)+b^{2} & \hspace{4cm} & & \text{Definition 2.3.11}\\ \\ & =a^{2}+(ab)+(ab)+b^{2} & \hspace{4cm} & & {a\times b=ab}\\ \\ & =a^{2}+(0+ab)+ab+b^{2} & \hspace{4cm} & & \text{Definition 2.2.1}\\ \\ & =a^{2}+(0\times ab+ab)+ab+b^{2} & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =a^{2}+((0++)\times ab)+ab+b^{2} & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =a^{2}+(1\times ab+ab)+b^{2} & \hspace{4cm} & & {0++=1}\\ \\ & =a^{2}+((1++)\times ab)+b^{2} & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =a^{2}+2ab+b^{2} & \hspace{4cm} & & {1++=2}\end{alignat*} $$
