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 by definition of a relation:
 * $\neg \exists s \in A: \left({s, t}\right) \in \mathcal R$

By Image of Subset is Subset of Image:
 * $\mathcal R \left[{A \setminus B}\right] \subseteq \mathcal R \left[{A}\right]$

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 Set Complement inverts 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.

That is, $s \ne t$.

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.