Condition for Operation to be Right Distributive over Constant Operation

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Theorem

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

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


Then:

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

if and only if:

$\forall x \in S: c \circ x = c$


Proof

Sufficient Condition

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

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

That is:

$\forall x \in S: c \circ x = c$

$\Box$


Necessary Condition

Let:

$\forall x \in S: c \circ x = c$

Then:

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

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

$\blacksquare$


Sources

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