Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation/Lemma

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Theorem

Let $S$ be a set.

Let $A \subsetneqq S$ be a proper subset of $S$.

Let $a \in A$ and $b \in \relcomp S A$.

Let $\odot$ be the operation on $S$ defined as:

$\forall x, y \in S: x \odot y = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}$


Then $\odot$ is commutative and associative.


Proof

Commutativity

Trivially:

\(\ds \forall x, y \in S: \, \) \(\ds y \odot x\) \(=\) \(\ds \begin {cases} a & : \set {y, x} \subseteq A \\ b & : \set {y, x} \not \subseteq A \end {cases}\) Definition of $\odot$
\(\ds \) \(=\) \(\ds \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}\) Definition of Set
\(\ds \) \(=\) \(\ds x \odot y\) Definition of Set

$\Box$


Associativity

There are $2$ cases to attend to:

Case $1$
$\set {x, y, z} \subseteq A$
\(\ds x \odot \paren {y \odot z}\) \(=\) \(\ds x \odot \paren {\begin {cases} a & : \set {y, z} \subseteq A \\ b & : \set {y, z} \not \subseteq A \end {cases} }\) Definition of $\odot$
\(\ds \) \(=\) \(\ds x \odot a\) as $\set {y, z} \subseteq A$
\(\ds \) \(=\) \(\ds \begin {cases} a & : \set {x, a} \subseteq A \\ b & : \set {x, a} \not \subseteq A \end {cases}\) Definition of Set
\(\ds \) \(=\) \(\ds a\) as $\set {x, a} \subseteq A$


\(\ds \paren {x \odot y} \odot z\) \(=\) \(\ds z \odot \paren {x \odot y}\) as $\odot$ is a priori commutative
\(\ds \) \(=\) \(\ds a\) from above
\(\ds \) \(=\) \(\ds x \odot \paren {y \odot z}\) from above

That is:

$x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = a$


Case $2$
$\set {x, y, z} \not \subseteq A$

Then at least one of $\set {x, y, z}$ is in $\relcomp S A$.

Without loss of generality, suppose $x \in \relcomp S A$.

Then:

$\forall y \in S: \set {x, y} \not \subseteq A$

and so:

$\forall y \in S: x \odot y = b$

Then we note that similarly:

$\forall y \in S: \set {b, y} \not \subseteq A$

and so:

$\forall y \in S: y \odot b = b$


Hence:

$x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = b$


Both cases have been examined, and we see that:

$\forall x, y, z \in S: x \odot \paren {y \odot z} = \paren {x \odot y} \odot z$

$\blacksquare$


Sources