How to find cardinality of $C \backslash (B \cup D)$? Let $B, C, D$ be finite sets $|B|=|C|  $ where $|\cdot|$ is the cardinality of the set.
Show that
$$
|C \backslash (B \cup D)|=|B \backslash C| -|(C \cap D) \backslash B|
$$
where $\backslash$ is the set difference.
 A: As pointed out by Zuy in the comment, I only have a proof for the case that $B$ and $C$ are finite sets:
By $|B|=|C|$, we have $|B\setminus C|=|C\setminus B|$.
Also, $C\setminus B$ is the disjoint union of $C\setminus (B\cup D)$ and $(C\cap D)\setminus B$. Thus, we have
$$|B\setminus C|=|C\setminus (B\cup D)|+|C\cap D)\setminus B|.$$
It has nothing to do with $|D|\leq |B|$.

In general, the equality can be disproved with $C=\mathbb{Z}$ and $B=D=\mathbb{Z}^+$:
We have

*

*$|B\setminus C|=|\mathbb{Z}^+\setminus \mathbb{Z}|=|\varnothing|=0$,

*$|C\setminus (B\cup D)|=|\mathbb{Z}\setminus \mathbb{Z}^+|=|\{0\}\cup \mathbb{Z}^-|=\aleph_0$, and

*$|(C\cap D)\setminus B|=|\mathbb{Z}^+\setminus \mathbb{Z}^+|=|\varnothing|=0$.

A: This is wrong in general. If $C$ is a strict subset of $B$ (for instance $B=\mathbb N$ and $C=\mathbb N_{>0}$), the left side equals $0$, while the right one is strictly greater than $0$.
A: The first step is to observe that the sets $C\setminus (B\cup D)$ and $(C\cap D)\setminus B$ form a partition of $C\setminus B$.
We know, then, that $\bigl|C\setminus (B\cup D)\bigr| = |C\setminus B| - \bigl|(C\cap D)\setminus B\bigr|$.
It suffices, therefore, to prove that $|C\setminus B| = |B\setminus C|$, which follows from $|B|=|C|$. The fact that $|D|\leq|B|$ seems useless.
