Mapping Images are Disjoint only if Domains are Disjoint

Theorem
Let $S$ and $T$ be sets.

Let:
 * $f \sqbrk S \cap f \sqbrk T = \O$

where $f \sqbrk S$ denotes the image set of $S$.

Then:
 * $S \cap T = \O$

Proof
From Image of Intersection under Mapping:
 * $f \sqbrk {S \cap T} \subseteq f \sqbrk S \cap f \sqbrk T$

From Empty Set is Subset of All Sets:
 * $f \sqbrk {S \cap T} = \O$

From Image of Subset under Mapping is Subset of Image:
 * $S \cap T = \O$