Given: $A \subset B$

To prove: $C - B \subset C - A$

Proof:

Let $x$ be an arbitrary element of $C - B$.

By definition of difference of sets, $x \in C - B$ means that $x \in C$ and $x \notin B$.

Since $A \subset B$ (given), every element of $A$ is also an element of $B$. This means if any element belongs to $A$, it must also belong to $B$.

Now, since $x \notin B$, and $A \subset B$, the element $x$ cannot belong to $A$. If $x$ were in $A$, then by the subset relation $A \subset B$, $x$ would have to be in $B$, which contradicts our condition that $x \notin B$.

Therefore, $x \notin A$.

So we have established that $x \in C$ and $x \notin A$. By definition of set difference, this means $x \in C - A$.

Since $x$ was an arbitrary element chosen from $C - B$, and we have shown that $x \in C - A$, we conclude that every element of $C - B$ is also an element of $C - A$.

Therefore, $C - B \subset C - A$.

Hence proved.