This question seems easy enough but I want to make sure

Q) Use the definition of a Boolean Algebra to give reasons for each step in the proof below:

$\displaystyle \forall a \in B, a \cdot a = a$

Proof:

Let a be any element of B, then

$\displaystyle \begin{array}{cc} a = a \cdot 1 \\=a \cdot (a + \bar{a})\\=(a\cdot a)+(a \cdot \bar{a} \\ =(a \cdot a) + 0 \\ =a \cdot a\end{array}$

first step. $\displaystyle a \cdot 1$ is like saying a and 1

second step $\displaystyle 1 = a + \bar{a}$ so we get $\displaystyle a \cdot (a + \bar{a})$

third step is distributive, the + is similar to an 'or' in logic

fourth step $\displaystyle a \cdot \bar{a} = 0$ similar to a set and its complement yield the empty set

fifth step is the left over $\displaystyle a \cdot a$ which is similar to a set intersect itself