Equivalence class

Mathematical statement

Given a complete finite carrier X and a certified equivalence relation R, define [a]={x in X:aRx}. If aRb then [a]=[b]; any two equivalence classes are equal or disjoint. Sampled relations or relations that fail the equivalence axioms must not be used to form canonical classes.

Representative formula (exact TeX)

K(R)

Assumptions

• The carrier 0 through n−1 and the complete relation-pair set are explicit and first pass full reflexivity, symmetry and transitivity checks.
• The class uses that same relation and carrier, not a visual guess at clustering.

Symbol scope

• K — equivalence class
• R — complete equivalence relation

Intuition, definition, and misconceptions

[a] lists exactly every element x for which aRx.

Related representatives produce the same class.

Classes are equal or disjoint and cover the carrier; quotient sets are a separate node.

An uncertified relation does not justify class formation, and two unequal equivalence classes cannot retain a nonempty overlap.

Adopted derivation guide

The original MKO is retained. The guide has its own append-only revision and verification boundaries, not a new Lean proof.

Prerequisites

• Equivalence relation — Classes use the same previously certified relation.
• Set membership — Class membership is equivalent to aRx.
• Subset — Each class is a subset of the carrier.

Verification boundary

Only the specified complete finite equivalence relation is handled; quotient operations and their independence from representative choice are not constructed.

not_formalized

This does not mean the content is false; it means no replayable computational or formal evidence is registered yet.

Next obligation

Add quotient sets and quotient rings of finite rings.

Same-source structured data

This page and the interactive view use the same object and existing language edition; this adds no proof or human review.