One-to-Many Image of Set Difference

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation.

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

Then:
 * $(1):\quad \mathcal R \left({A}\right) \setminus \mathcal R \left({B}\right) = \mathcal R \left({A \setminus B}\right)$

iff $\mathcal R$ is one-to-many.

Sufficient Condition
First, to show that $(1)$ holds if $\mathcal R$ is one-to-many.

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 $t \notin \mathcal R \left({A}\right) \lor t \in \mathcal R \left({B}\right)$ 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: $S \subseteq T \iff \complement \left({T}\right) \subseteq \complement \left({S}\right)$.

Necessary Condition
Now for the converse: If $(1)$ holds, it is to be shown that $\mathcal R$ is one-to-many.

Let $s,t \in S$ be distinct, i.e., $s \ne t$.

Then in particular $\left\{{s}\right\} \setminus \left\{{t}\right\} = \left\{{s}\right\}$.

Applying $(1)$ to these two sets, it follows that:


 * $\mathcal R \left({\left\{{s}\right\}}\right) \setminus \mathcal R \left({\left\{{t}\right\}}\right) = \mathcal R \left({\left\{{s}\right\}}\right)$

By Set Difference with Disjoint Set, this implies that:


 * $\mathcal R \left({\left\{{s}\right\}}\right) \cap \mathcal R \left({\left\{{t}\right\}}\right) = \varnothing$

It follows that every element of $T$ can be related to at most one element of $S$.

That is, $\mathcal R$ is one-to-many.