I am reading G.B. Fine's book, "The Number System Of Algebra". He defines the number $a-b$ as the solution of the equation $a=x+b$, or the number satisfying the equation $(a-b)+b=a$.
In article 13 he claims that $a-a$ has value which is independent of the value of $a$ and acts as an additive identity. He proves it as,
$(a-a)+(a+b)=(((a-a)+a)+b)$
by definition $((a-a)+a)$ is $a$. So the above written equation becomes,
$(a-a)+(a+b)=(a+b)$
Now, $(b-b)+(a+b)=(((b-b)+b)+a)))=b+a=a+b$
So we see that any number of form $a-a$ is altogether independent of $a$ and may properly be represented by a symbol unrelated to $a$. The symbol which has been chosen for it is $0$.
- The problem is that how can we show that the quantity $(a-a)+s$ is equal to the quantity $(b-b)+s$, i.e. both the quantiteis are independnt of $a$ and $b$.
I just overlooked the article. It is easy to prove that $a-a=b-b$. It is due to the axiom-VII:
If $$\begin{cases} \text{If}\ \ a+c=b+c \\ \ \ \ \ \ \ \ \ \ \ \ \ a=b \end{cases}$$
We have $$(a−a)+(a+b)=(b-b)+(a+b)$$ So, $(a-a)=(b-b)$