Domain of Composite Relation

Theorem
Let $$\mathcal R_2 \circ \mathcal R_1$$ be a composite relation.

Then the domain of $$\mathcal R_2 \circ \mathcal R_1$$ is the domain of $$\mathcal R_1$$:


 * $$\operatorname{Dom} \left ({\mathcal R_2 \circ \mathcal R_1}\right) = \operatorname{Dom} \left ({\mathcal R_1}\right)$$

Proof
Let $$\mathcal R_1 \subseteq S_1 \times S_2$$ and $$\mathcal R_2 \subseteq S_2 \times S_3$$.

The domain of $$\mathcal R_1$$ is $$S_1$$.

The composite of $$\mathcal R_1$$ and $$\mathcal R_2$$ is defined as:


 * $$\mathcal R_2 \circ \mathcal R_1 = \left\{{\left({x, z}\right): x \in S_1, z \in S_3: \exists y \in S_2: \left({x, y}\right) \in \mathcal R_1 \and \left({y, z}\right) \in \mathcal R_2}\right\}$$

From this definition:
 * $$\mathcal R_2 \circ \mathcal R_1 \subseteq S_1 \times S_3$$.

Thus the domain of $$\mathcal R_2 \circ \mathcal R_1$$ is $$S_1$$.

Thus:
 * $$\operatorname{Dom} \left ({\mathcal R_2 \circ \mathcal R_1}\right) = S_1 = \operatorname{Dom} \left ({\mathcal R_1}\right)$$