Set Difference with Proper Subset is Proper Subset

Theorem
Let $S$ be a set.

Let $T \subsetneq S$ be a proper subset of $S$. Let $S \setminus T$ denote the set difference between $S$ and $T$.

Then:


 * $S \setminus T$ is a proper subset of $S$

Proof
From Set Difference is Subset:
 * $S \setminus T \subseteq S$

From Set Difference with Proper Subset:
 * $S \setminus T \ne \O$

By definition of proper subset:
 * $T \ne \O$

From Intersection with Subset is Subset:
 * $S \cap T = T$

Hence:
 * $S \cap T \ne \O$

From the contrapositive statement of Set Difference with Disjoint Set:
 * $S \setminus T \ne S$

It follows that $S \setminus T$ is a proper subset by definition.