Set Difference Equals First Set iff Empty Intersection

Theorem

 * $$S - T = S \iff S \cap T = \varnothing$$

Proof
Suppose $$S - T = S$$. Then, by definition of set equality, we have $$S \subseteq S - T$$. Thus:

$$ $$ $$ $$ $$ $$ $$ $$

Thus we have that $$S - T = S \implies S \cap T = \varnothing$$.

Similarly, we have $$S \cap T = \varnothing \implies S \subseteq S - T$$ from the above.

But from Set Difference Subset we also have $$S - T \subseteq S$$.

Thus by the definition of set equality, it follows that $$S \cap T = \varnothing \implies S - T = S$$.

The implications therefore go both ways and the proof is complete.