One-to-Many Image of Set Difference

Theorem
Let $$\mathcal{R} \subseteq S \times T$$ be a relation which is one-to-many.

Let $$A$$ and $$B$$ be subsets of $$S$$.

Then:
 * $$\mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right) = \mathcal{R} \left({A \setminus B}\right)$$

Proof
From Image of Set Difference, we already have:


 * $$\mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right) \subseteq \mathcal{R} \left({A \setminus B}\right)$$

So we just need to show:


 * $$\mathcal{R} \left({A \setminus B}\right) \subseteq \mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right)$$

Let $$t \notin \mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right)$$.

Then $$t \notin \mathcal{R} \left({A}\right) \or t \in \mathcal{R} \left({B}\right)$$ by De Morgan's Laws.


 * Suppose $$t \notin \mathcal{R} \left({A}\right)$$. Then $$\lnot \exists s \in A: \left({s, t}\right) \in \mathcal{R}$$ by definition of a relation.

But $$\mathcal{R} \left({A \setminus B}\right) \subseteq \mathcal{R} \left({A}\right)$$ by Subset of Image.

Thus, by definition of subset and Rule of Transposition, $$t \notin \mathcal{R} \left({A}\right) \implies t \notin \mathcal{R} \left({A \setminus B}\right)$$.


 * Now suppose $$t \in \mathcal{R} \left({B}\right)$$.

Then $$\exists s \in B: \left({s, t}\right) \in \mathcal{R}$$.

Because $$\mathcal{R}$$ is one-to-many, $$\forall x \in S: \left({x, t}\right) \in \mathcal{R} \implies x = s$$ and thus $$x \in B$$.

Thus $$x \notin A \setminus B$$ and hence $$t \notin \mathcal{R} \left({A \setminus B}\right)$$.


 * So by Proof by Cases, $$t \notin \mathcal{R} \left({A}\right) \setminus \mathcal{R} \left({B}\right) \implies t \notin \mathcal{R} \left({A \setminus B}\right)$$.

The result follows from Complements Invert Subsets: $$S \subseteq T \iff \mathcal{C} \left({T}\right) \subseteq \mathcal{C} \left({S}\right)$$.