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 $\mathcal R_1$:


 * $\Dom {\mathcal R_2 \circ \mathcal R_1} = \Dom {\mathcal R_1}$

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:


 * $\mathcal R_2 \circ \mathcal R_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \mathcal R_1 \land \tuple {y, z} \in \mathcal R_2}$

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:
 * $\Dom {\mathcal R_2 \circ \mathcal R_1} = S_1 = \Dom {\mathcal R_1}$