# Set Difference with Proper Subset

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## 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:

- $S \setminus T \ne \O$

where $\O$ denotes the empty set.

## Proof

Aiming for a contradiction, suppose $S \setminus T = \O$.

Then:

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

By De Morgan's laws:

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

By definition of subset:

- $S \subseteq T$

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

But we have $T \subseteq S$ and $S \subseteq T$.

So by definition of set equality:

- $S = T$

From this contradiction it follows that:

- $S \setminus T \ne \O$

$\blacksquare$