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 $\RR$ be an equivalence relation on $S$.

Every equivalence relation is reflexive, so:

$c \mathrel \RR c$

So:

\(\ds x_1 \mathrel \RR x_2\)

\(\land\)

\(\ds y_1 \mathrel \RR y_2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {x_1 \sqbrk c y_1}\)

\(\RR\)

\(\ds \paren {x_2 \sqbrk c y_2}\)

True Statement is implied by Every Statement

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.4$