Show $\left|B-A\right| = \left|B'-A'\right|$

Question

Let $A,A',B,B'$ such that: $A\subseteq B$, $A'\subseteq B'$, $\left|B\right|=\left|B'\right| \gt \aleph_0$, $\left|A\right|=\left|A'\right| = \aleph_0$.
Show that $\left|B-A\right| = \left|B'-A'\right|$

As a start I know that $\left|B-A\right| > \aleph_0$. Otherwise,
$$\left|B\right| = \left|B-A\right| + \left|A\right| \le \aleph_0 + \aleph_0 \le \aleph_0$$

In contrary to $\left|B\right| > \aleph_0$.

For same reason, $\left|B'-A'\right| > \aleph_0$.

How to proceed? I'm kinda stuck at this point.

Answer 1

Taking up on Arthur's suggestion above, we prove that $|B| = |B\setminus A|$: Let $\kappa = |B\setminus A|$ and $\lambda = |B\cap A|$. Since $B\setminus A$ and $B\cap A$ are disjoint sets whose union is $B$ we get that $\kappa + \lambda = |B|$. The cardinal $\kappa$ is infinite, so $\kappa + \lambda = \max\{\kappa, \lambda\} = \kappa$ (because $\lambda$ is at most $\aleph_0$). So $|B| = \kappa = |B\setminus A|$.

Now the same argument for $A', B'$ instead of $A,B$ gives $|B'| = \kappa = |B'\setminus A'|$ because $|B'| = |B| = \kappa$. So $|B'\setminus A'| = |B\setminus A|$.

Answer 2

$|B|>\aleph_0$ and $B=(B \setminus A) \cup A$. Thus $|B| = \max\{|B \setminus A|,|A|\} = |B \setminus A|$ because of $|A|=\aleph_0$.

Likewise $|B'|=|B' \setminus A'|$.