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

Corollary 1
In addition to the other conditions above:

Let $A \subseteq B$.

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

where $\complement$ (in this context) denotes relative complement.

Corollary 2

 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({\complement_S \left({A}\right)}\right)$

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

Proof of Corollary 1
We have that $A \subseteq B$.

Then by definition of relative complement:
 * $\complement_B \left({A}\right) = B \setminus A$
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right)$

Hence, when $A \subseteq B$:
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({\complement_B \left({A}\right)}\right)$

means exactly the same thing as:
 * $\mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right) = \mathcal R \left({B \setminus A}\right)$

Proof of Corollary 2
By definition of the image of $\mathcal R$:
 * $\operatorname{Im} \left({\mathcal R}\right) = \mathcal R \left({S}\right)$

So, when $B = S$ in Corollary 1:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \complement_{\mathcal R \left({S}\right)} \left({\mathcal R \left({A}\right)}\right)$

Hence:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({\complement_S \left({A}\right)}\right)$

means exactly the same thing as:
 * $\complement_{\mathcal R \left({S}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({\complement_S \left({A}\right)}\right)$

that is:
 * $\mathcal R \left({S}\right) \setminus \mathcal R \left({A}\right) = \mathcal R \left({S \setminus A}\right)$