Inverse of Composite Relation

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Theorem

Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the composite of the two relations $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.


Then:

$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$


Proof

Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be relations.

We assume that:

$\Dom {\RR_2} = \Cdm {\RR_1}$

where $\Dom \RR$ denotes domain and $\Cdm \RR$ denotes codomain of a relation $\RR$.

This is necessary for $\RR_2 \circ \RR_1$ to exist.


From the definition of an inverse relation, we have:

$\Dom {\RR_2} = \Cdm {\RR_2^{-1} }$
$\Cdm {\RR_1} = \Dom {\RR_1^{-1} }$


So we confirm that $\RR_1^{-1} \circ \RR_2^{-1}$ is defined.


\(\displaystyle \RR_2 \circ \RR_1\) \(=\) \(\displaystyle \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}\) Definition of Composition of Relations
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\RR_2 \circ \RR_1}^{-1}\) \(=\) \(\displaystyle \set {\tuple {z, x}: \tuple {x, z} \in \RR_2 \circ \RR_1}\) Definition of Inverse Relation
\(\displaystyle \) \(=\) \(\displaystyle \set {\tuple {z, x}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}\) Definition of Composition of Relations
\(\displaystyle \) \(=\) \(\displaystyle \set {\tuple {z, x}: z \in S_3, x \in S_1: \exists y \in S_2: \tuple {z, y} \in \RR_2^{-1} \land \tuple {y, x} \in \RR_1^{-1} }\) Definition of Inverse Relation
\(\displaystyle \) \(=\) \(\displaystyle \RR_1^{-1} \circ \RR_2^{-1}\) Definition of Composition of Relations

$\blacksquare$


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