Proof of commutativity and associativity for binary error pattern

There's a math problem on page-34 of Pinter's "A Book of Abstract Algebra" book which is(Problem F):

If a word is $\mathbf{a} = a_1 a_2 \cdots a_n$ is sent, but a word $\mathbf{b} = b_1 b_2 \cdots b_n$ is received(where the $a_i$
and $b_j$ are $0s$ or $1s$ ), then the error pattern is the word $e = e_1 e_2 ... e_n$ where:

$e_i = \begin{cases} 0 & \text{ if } a_i = b_i \\ 1 & \text{ if } a_i \neq b_i \end{cases}$

With this motivation, we define an operation of adding words, as follows: If $\mathbf{a}$ and $\mathbf{b}$ are both of length
$1$ , we add them according to the rules

$0 + 0 = 0$ , $1 + 1 = 0$ , $0 + 1 = 1$ and $1 + 0 = 1$

If $\mathbf{a}$ and $\mathbf{b}$ are both of length $n$ , we add them by adding corresponding digits. That is(let us
introduce commas for convenience),

$(a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots, b_n) = (a_1 + b_1, a_2 + b_2, \cdots, a_n + b_n)$

Thus the sum of $\mathbf{a}$ and $\mathbf{b}$ is the error pattern $\mathbf{e}$ .

For example,

$0010110 + 0011010 = 0001100$ and $10100111 + 11110111 = 01010000$

The symbol $\mathbb{B}^n$ will designate the set of all the binary words of length $n$ .

We will prove that the operation of word addition has the following properties on $\mathbb{B}^n$ :

It is commutative.
It is associative.
There is an identity element for word addition.
Every word has an inverse under word addition.

.....

1) Show that

$(a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots b_n) = (b_1, b_2, \cdots, b_n) + (a_1, a_2, \cdots, a_n)$

3) Show that

$\begin{align*}(a_1, a_2, \cdots, a_n) + [(b_1, b_2, \cdots, b_n) + (c_1, c_2, \cdots, c_n)] =& [ (a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots, b_n) ] \\ + & (c_1, c_2, \cdots, c_n) \end{align*}$ .

How can I prove that the commutativity and associativity holds in this case?