Image of Set Difference under Mapping/Proof 1

Proof
As $f$, being a mapping, is also a relation, we can apply Image of Set Difference under Relation:


 * $\mathcal R \sqbrk {S_1} \setminus \mathcal R \sqbrk {S_2} \subseteq \mathcal R \sqbrk {S_1 \setminus S_2}$