# Equivalence Relation is Congruence for Constant Operation

## Theorem

Every equivalence relation is a congruence relation for the constant operation.

## Proof

Let $c \in S$.

By the definition of the constant operation, $\forall x, y \in S: x \sqbrk c y = c$.

Let $\mathcal R$ be an equivalence relation on $S$.

Every equivalence relation is reflexive, so:

- $c \mathop {\mathcal R} c$

So:

\(\displaystyle x_1 \mathop {\mathcal R} x_2\) | \(\land\) | \(\displaystyle y_1 \mathop {\mathcal R} y_2\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {x_1 \sqbrk c y_1}\) | \(\mathcal R\) | \(\displaystyle \paren {x_2 \sqbrk c y_2}\) | True Statement is implied by Every Statement |

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 11$: Example $11.4$