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

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