If a word $a = a_1,a_2,...,a_n$ is sent is sent (this is in regards to coding/IT/etc.-I'm trying not to include any extraneous information) and a word $b= b_1,b_2,...,b_n$ is received (where the $a_{i}$ and the $b_{j}$ are $0$s or $1$s), then the error pattern is the word $e = e_1,e_2,...e_n$ where
$e_{i} = 0$ if $a_{i} = b_{i}$ and $e_{i} = 1$ if $a_{i} \neq b_{i}$ .
With this motivation, we define an operation of adding words as follows: if a and b are both of length 1, we add them as according to the rules
$0 + 0 = 0$ $1 + 1 = 0$ $0 + 1 = 1 $1 + 0 = 1$
If a and b are both of length n, we add them by adding corresponding digits. That is
$(a_1, a_2,..., a_n) + (b_1, b_2,..., b_n) = (a_1+ b_1, a_2 + b_2,..., a_n + b_n)$
So the sum of a and b is the error pattern, e.
I have shown that this operation is commutative, associative, that the identity element for adding words of length n is $0_n$ and that the inverse of any word $(a_1,...,a_n)$ in the set $\Bbb B^*n*$ (the set of all binary words of length n is $(a_1,...,a_n)$.
Now I'm looking to show that
$a + b = a - b$ where $a - b= a + (-b)$
And that
if $a+ b = c$ then $a = b+c$
Your plus and minus are bitwise XOR. In $\Bbb F_2$ (which you are working in) negation is the identity because $-1 = 1$ The things you want to show follow from that.