(related to Proposition: Addition of Integers)
Let \(x\) and \(y\) be integers. By definition, it means that they are equivalence classes represented by some natural numbers \(x=\lbrack a,b\rbrack \), \(y=\lbrack c,d\rbrack \), \(a,b,c,d\in \mathbb N\).
Note that \(a+b\), \(c+d\) are all natural numbers, since they are the sums of the respective natural numbers \(a,c\) and \(b,d\). Therefore, the sum $$\begin{array}{rcl} x+y=\lbrack a+c,b+d\rbrack . \end{array}$$ exists, because it denotes some new integer, as it is represented by the natural numbers \(a+c\) and \(b+d\).
It remains to be shown that the addition of integers does not depend on the specific representatives of the numbers \(x\) and \(y\). Suppose, we have different representatives $$\begin{array}{rcl} x=\lbrack a_1,b_1\rbrack =\lbrack a_2,b_2\rbrack ,~y=\lbrack c_1,d_1\rbrack =\lbrack c_2,d_2\rbrack .&&(*) \end{array}$$ Without loss of generality, we can assume \(a_1\ge a_2\) and \(c_1\ge c_2\). It follows from the definition of integers that there exist some natural numbers \(i,j\) with $$\begin{array}{rl} a_1=a_2+i,&c_1=c_2+j,\\ b_1=b_2+i,&d_1=d_2+j.\\ \end{array}\quad (*)$$
We have to show that $$x+y=\lbrack a_1+c_1,b_1+d_1\rbrack =\lbrack a_2+c_2,b_2+d_2\rbrack .$$ In the following, we will use the following mathematical definitions and concepts: * definition of integers, * definition of adding integers (hypothesis), * commutativity law of adding natural numbers, * cancellation law of adding natural numbers, and * associativity law of adding natural numbers:
$$\begin{array}{rcll} x+y&=&\lbrack a_1,b_1\rbrack +\lbrack c_1,d_1\rbrack &\text{by definition of integers}\\ &=&\lbrack a_1+c_1,b_1+d_1\rbrack &\text{by hypothesis}\\ &=&\lbrack (a_2+i)+(c_2+j),(b_2+i)+(d_2+j)\rbrack &\text{according to }(*)\\ &=&\lbrack a_2+i+c_2+j,b_2+i+d_2+j\rbrack &\text{by associativity law for adding natural numbers}\\ &=&\lbrack a_2+c_2+i+j,b_2+d_2+i+j\rbrack &\text{by commutativity law for adding natural numbers}\\ &=&\lbrack a_2+c_2+(i+j),b_2+d_2+(i+j)\rbrack &\text{by associativity law for adding natural numbers}\\ &=&\lbrack a_2+c_2+\cancel{(i+j)},b_2+d_2+\cancel{(i+j)}\rbrack &\text{by cancellation law for adding natural numbers}\\ &=&\lbrack a_2+c_2,b_2+d_2\rbrack &\text{by definition of integers}\\ &=&\lbrack a_2,b_2\rbrack +\lbrack c_2,d_2\rbrack &\text{by hypothesis}\\ \end{array}$$
