Properties of integer halving by powers of 2. I'm dividing integers with a method where the remainder is removed after division. 
Example: 17/2 = 8 , 11/2 = 5 , 8/2 = 4 , 2/2 =  1
Is this property true for every integer; dividing by 4 yields the same result as dividing two times by 2 or dividing by 8 yields the same result as dividing three times by 2 and so on.

This process is called integer division. Your quesion is then:
For $n \in \mathbb N$ is it true that $\def\b{\backslash}$
$$n \b 4 = (n \b 2) \b 2$$
Where $\b$ denotes integer division. To prove this, we consider only $n = 4k + i, n\b 4=k$ where $i\in \{1,2,3\}$, since the remainig case is trivial.
1. $i = 1$:
$$((4k + 1) \b 2) \b 2 = 2k \b 2 = k$$
2. $i = 2$:
$$((4k + 2) \b 2) \b 2 = (2k + 1)\b2 = k$$
3. $i=3$:
$$((4k + 3) \b 2) \b 2 = (2k + 1) \b 2 = k$$
This concludes the proof.
