Set Difference with Subset is Superset of Set Difference

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Theorem

Let $A, B, S$ be sets or classes.

Suppose that $A \subseteq B$.


Then $S \setminus B \subseteq S \setminus A$, where $\setminus$ represents set difference.


Proof

Let $x \in S \setminus B$.

Then by the definition of set difference:

$x \in S$ and $x \notin B$

Suppose for the sake of contradiction that $x \in A$.

Then since $A$ is a subset (or subclass) of $B$, $x \in B$, a contradiction.

Thus $x \notin A$.

Since $x \in S$ and $x \notin A$, we conclude that $x \in S \setminus A$.

As this holds for all $x \in S \setminus B$:

$S \setminus B \subseteq S \setminus A$

$\blacksquare$