Restriction of Commutative Operation is Commutative

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Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $T \subseteq S$.

Let the operation $\circ$ be commutative on $\struct {S, \circ}$.


Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative.


Proof

\(\ds T\) \(\subseteq\) \(\ds S\)
\(\ds \leadsto \ \ \) \(\ds \forall a, b \in T: \, \) \(\ds a, b\) \(\in\) \(\ds S\) Definition of Subset
\(\ds \leadsto \ \ \) \(\ds a \mathop {\circ {\restriction_T} } b\) \(=\) \(\ds a \circ b\)
\(\ds \) \(=\) \(\ds b \circ a\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds b \mathop {\circ {\restriction_T} } a\)

$\blacksquare$


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