To prove that:

$|C| = |C \cap A| + |(C \cap B) \setminus A| + |C \setminus B|$

given that $A \subseteq B$, we will proceed by examining each part of the equation separately and understanding what each term represents in terms of set elements.

Step-by-Step Proof

1. Breaking down the sets:
   We are given three terms on the right side of the equation:

   • $|C \cap A|$: the number of elements in both $C$ and $A$.
   • $|(C \cap B) \setminus A|$: the number of elements in $C \cap B$ that are not in $A$.
   • $|C \setminus B|$: the number of elements in $C$ that are not in $B$.

   These three terms are mutually exclusive, meaning each element in $C$ falls into one of these three categories without overlap.

2. Partitioning of $C$: Since $A \subseteq B$, any element of $C$ can belong to one of the following mutually exclusive subsets:

   • $C \cap A$: elements in both $C$ and $A$ (and therefore also in $B$, since $A \subseteq B$).
   • $(C \cap B) \setminus A$: elements in $C$ and $B$ but not in $A$.
   • $C \setminus B$: elements in $C$ that are not in $B$ (and hence also not in $A$, since $A \subseteq B$).

   Thus, every element of $C$ must belong to one of these three subsets.

3. Counting the elements of $C$: Since these subsets are mutually exclusive and cover all elements of $C$, we can add the cardinalities of each subset to find the total number of elements in $C$:

   $|C| = |C \cap A| + |(C \cap B) \setminus A| + |C \setminus B|$

4. Conclusion: This completes the proof, as we have shown that the elements in $C$ are partitioned into three mutually exclusive parts, whose cardinalities sum to $|C|$.

Summary

Thus, we have proven that if $A \subseteq B$, then:

$|C| = |C \cap A| + |(C \cap B) \setminus A| + |C \setminus B|$

This follows by partitioning $C$ into three distinct subsets and adding their cardinalities.

Would you like more details on any specific part of this proof?

---