Condition for Constant Operation to be Distributive over Another Operation

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Theorem

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

Let $\sqbrk c$ denote the constant operation for some $c \in S$.


Then:

$\sqbrk c$ is distributive over $\circ$

if and only if:

$c \circ c = c$


Proof

Sufficient Condition

Let $\sqbrk c$ be distributive over $\circ$

\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \sqbrk c y} \circ \paren {x \sqbrk c z}\) \(=\) \(\ds c circ c\) Definition of Constant Operation
\(\ds \forall x, y, z \in S: \, \) \(\ds x \sqbrk c \paren {y \circ z}\) \(=\) \(\ds c\) Definition of Constant Operation
$c \circ c = c$

$\Box$


Necessary Condition

Let $\forall c \circ c = c$.

Then:

\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \sqbrk c y} \circ \paren {x \sqbrk c z}\) \(=\) \(\ds c \circ c\) Definition of Constant Operation
\(\ds \) \(=\) \(\ds c\) by hypothesis
\(\ds \) \(=\) \(\ds x \sqbrk c \paren {y \circ z}\) Definition of Constant Operation

and:

\(\ds \forall x, y, z \in S: \, \) \(\ds \paren {x \sqbrk c z} \circ \paren {y \sqbrk c z}\) \(=\) \(\ds c \circ c\) Definition of Constant Operation
\(\ds \) \(=\) \(\ds c\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {x \circ y} \sqbrk c z\) Definition of Constant Operation

That is, $\sqbrk c$ is distributive over $\circ$.

$\blacksquare$


Sources

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