Equivalence Relation is Congruence for Constant Operation

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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