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 $$\mathcal{R}_1$$:

$$\mathrm {Dom} \left ({\mathcal{R}_2 \circ \mathcal{R}_1}\right) = \mathrm {Dom} \left ({\mathcal{R}_1}\right)$$

Proof
Let $$\mathcal{R}_1 \subseteq S_1 \times S_2$$ and $$\mathcal{R}_2 \subseteq S_2 \times S_3$$.

The composite of $$\mathcal{R}_1$$ and $$\mathcal{R}_2$$ is defined as:

$$\mathcal{R}_2 \circ \mathcal{R}_1 = \left\{{\left({x, z}\right): x \in S_1, z \in S_3: \exists y \in S_2: \left({x, y}\right) \in \mathcal{R}_1 \land \left({y, z}\right) \in \mathcal{R}_2}\right\}$$

From this definition: $$\mathcal{R}_2 \circ \mathcal{R}_1 \subseteq S_1 \times S_3$$.

Thus $$\mathrm {Dom} \left ({\mathcal{R}_2 \circ \mathcal{R}_1}\right) = S_1 = \mathrm {Dom} \left ({\mathcal{R}_1}\right)$$.