Set Difference with Proper Subset

Theorem
Let $S$ be a non-empty set.

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

Then:


 * $S \setminus T \ne \varnothing$

where $\varnothing$ denotes the empty set.

Proof
Suppose $S \setminus T = \varnothing$.

That means:


 * $\not \exists x \in S: x \notin T$

By De Morgan's laws, that means:


 * $\forall x \in S: x \in T$

By definition of subset, that means:


 * $S \subseteq T$

By definition of proper subset, we have that $T \subseteq S$ such that $T \ne S$.

But as we have $T \subseteq S$ and $S \subseteq T$, it follows from Equality of Sets that $S = T$.

From this contradiction it follows that:


 * $S \setminus T \ne \varnothing$