# Complete Classes, Membership, Conditional Representative Invariance and Disjointness

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261029-r07-equivalence-class-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0. Give (element,class).

View answer and explanation(0, [0, 1, 2, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (0, [0, 1, 2, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0001/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 2. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1. Give (element,class).

View answer and explanation(1, [1, 2, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (1, [1, 2, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0002/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 3. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2. Give (element,class).

View answer and explanation(2, [0, 2, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (2, [0, 2, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0003/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 4. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3. Give (element,class).

View answer and explanation(3, [2, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (3, [2, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0004/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 5. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4. Give (element,class).

View answer and explanation(4, [2, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (4, [2, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0005/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 6. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 0], [3, 1], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5. Give (element,class).

View answer and explanation(5, [0, 1, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (5, [0, 1, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0006/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 7. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [1, 3], [1, 4], [1, 5], [2, 0], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0. Give (element,class).

View answer and explanation(0, [0, 2])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (0, [0, 2]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0007/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 8. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1. Give (element,class).

View answer and explanation(1, [1, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (1, [1, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0008/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 9. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [1, 2], [2, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2. Give (element,class).

View answer and explanation(2, [1, 2])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (2, [1, 2]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0009/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 10. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3. Give (element,class).

View answer and explanation(3, [3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (3, [3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0010/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 11. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4. Give (element,class).

View answer and explanation(4, [3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (4, [3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0011/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 12. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5. Give (element,class).

View answer and explanation(5, [0, 3, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (5, [0, 3, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0012/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 13. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [2, 0], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0. Give (element,class).

View answer and explanation(0, [0, 2])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (0, [0, 2]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0013/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 14. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1. Give (element,class).

View answer and explanation(1, [0, 1])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (1, [0, 1]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0014/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 15. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 1], [4, 2], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2. Give (element,class).

View answer and explanation(2, [0, 1, 2, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (2, [0, 1, 2, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0015/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 16. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3. Give (element,class).

View answer and explanation(3, [0, 3])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (3, [0, 3]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0016/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 17. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4. Give (element,class).

View answer and explanation(4, [1, 2, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (4, [1, 2, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0017/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 18. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [1, 3], [2, 0], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5. Give (element,class).

View answer and explanation(5, [0, 2, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (5, [0, 2, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0018/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 19. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [1, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 0], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0. Give (element,class).

View answer and explanation(0, [0, 1, 3])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (0, [0, 1, 3]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0019/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 20. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1. Give (element,class).

View answer and explanation(1, [1, 3])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (1, [1, 3]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0020/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 21. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2. Give (element,class).

View answer and explanation(2, [0, 2, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (2, [0, 2, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0021/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 22. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3. Give (element,class).

View answer and explanation(3, [0, 3])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (3, [0, 3]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0022/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 23. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4. Give (element,class).

View answer and explanation(4, [2, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (4, [2, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0023/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 24. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 4], [0, 5], [1, 0], [1, 1], [1, 4], [1, 5], [2, 2], [2, 3], [3, 2], [3, 3], [4, 0], [4, 1], [4, 4], [4, 5], [5, 0], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5. Give (element,class).

View answer and explanation(5, [0, 1, 4, 5])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (5, [0, 1, 4, 5]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0024/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 25. An Equivalence Class Contains Every Related Carrier Element

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [1, 1], [1, 4], [1, 5], [2, 0], [2, 2], [2, 3], [3, 0], [3, 2], [3, 3], [4, 1], [4, 4], [4, 5], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0. Give (element,class).

View answer and explanation(0, [0, 2, 3])

Filter the entire carrier by the certified relation with this representative. Keep every related element, including the representative by reflexivity. The exact result, retaining the requested order, is (0, [0, 2, 3]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0025/)

Equivalence class · qf-mod29-eqclass-of · candidate





## 26. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(0, 1, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (0, 1, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0026/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 27. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(1, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (1, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0027/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 28. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(2, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (2, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0028/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 29. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(3, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (3, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0029/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 30. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(4, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (4, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0030/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 31. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 0], [3, 1], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(5, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (5, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0031/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 32. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [1, 3], [1, 4], [1, 5], [2, 0], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(0, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (0, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0032/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 33. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(1, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (1, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0033/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 34. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [1, 2], [2, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(2, 1, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (2, 1, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0034/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 35. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(3, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (3, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0035/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 36. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(4, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (4, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0036/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 37. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(5, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (5, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0037/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 38. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [2, 0], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(0, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (0, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0038/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 39. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(1, 4, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (1, 4, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0039/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 40. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 1], [4, 2], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(2, 1, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (2, 1, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0040/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 41. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(3, 4, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (3, 4, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0041/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 42. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(4, 1, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (4, 1, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0042/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 43. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [1, 3], [2, 0], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(5, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (5, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0043/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 44. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [1, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 0], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(0, 1, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (0, 1, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0044/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 45. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=1, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(1, 4, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (1, 4, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0045/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 46. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=2, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(2, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (2, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0046/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 47. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=3, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(3, 4, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (3, 4, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0047/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 48. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=4, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(4, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (4, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0048/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 49. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 4], [0, 5], [1, 0], [1, 1], [1, 4], [1, 5], [2, 2], [2, 3], [3, 2], [3, 3], [4, 0], [4, 1], [4, 4], [4, 5], [5, 0], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=5, candidate=4. Give (element,candidate,related?,in_class?).

View answer and explanation(5, 4, Yes, Yes)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (5, 4, Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0049/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 50. Relation Membership and Class Membership Must Agree

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [1, 1], [1, 4], [1, 5], [2, 0], [2, 2], [2, 3], [3, 0], [3, 2], [3, 3], [4, 1], [4, 4], [4, 5], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. element=0, candidate=1. Give (element,candidate,related?,in_class?).

View answer and explanation(0, 1, No, No)

Look up the ordered relation pair and independently read candidate membership in the full class. Both flags use the same carrier and relation. The exact result, retaining the requested order, is (0, 1, No, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0050/)

Equivalence class · qf-mod29-eqclass-membership · candidate





## 51. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(0, 1, Yes, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (0, 1, Yes, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0051/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 52. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(1, 4, Yes, [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (1, 4, Yes, [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0052/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 53. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(2, 1, No, [0, 2, 3, 4, 5], [1], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (2, 1, No, [0, 2, 3, 4, 5], [1], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0053/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 54. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(3, 4, Yes, [2, 3, 4, 5], [2, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (3, 4, Yes, [2, 3, 4, 5], [2, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0054/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 55. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(4, 1, No, [2, 3, 4, 5], [1], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (4, 1, No, [2, 3, 4, 5], [1], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0055/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 56. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 0], [3, 1], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(5, 4, Yes, [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (5, 4, Yes, [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0056/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 57. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [1, 3], [1, 4], [1, 5], [2, 0], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(0, 1, No, [0, 2], [1, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (0, 1, No, [0, 2], [1, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0057/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 58. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(1, 4, Yes, [1, 3, 4, 5], [1, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (1, 4, Yes, [1, 3, 4, 5], [1, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0058/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 59. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [1, 2], [2, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(2, 1, Yes, [1, 2], [1, 2], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (2, 1, Yes, [1, 2], [1, 2], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0059/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 60. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(3, 4, Yes, [3, 4, 5], [3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (3, 4, Yes, [3, 4, 5], [3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0060/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 61. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(4, 1, No, [3, 4, 5], [1, 2], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (4, 1, No, [3, 4, 5], [1, 2], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0061/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 62. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(5, 4, Yes, [0, 3, 4, 5], [0, 3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (5, 4, Yes, [0, 3, 4, 5], [0, 3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0062/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 63. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [2, 0], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(0, 1, No, [0, 2], [1], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (0, 1, No, [0, 2], [1], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0063/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 64. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(1, 4, No, [0, 1], [3, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (1, 4, No, [0, 1], [3, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0064/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 65. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 1], [4, 2], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(2, 1, Yes, [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (2, 1, Yes, [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0065/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 66. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(3, 4, No, [0, 3], [1, 2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (3, 4, No, [0, 3], [1, 2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0066/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 67. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(4, 1, Yes, [1, 2, 4, 5], [1, 2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (4, 1, Yes, [1, 2, 4, 5], [1, 2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0067/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 68. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [1, 3], [2, 0], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(5, 4, Yes, [0, 2, 4, 5], [0, 2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (5, 4, Yes, [0, 2, 4, 5], [0, 2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0068/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 69. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [1, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 0], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(0, 1, Yes, [0, 1, 3], [0, 1, 3], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (0, 1, Yes, [0, 1, 3], [0, 1, 3], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0069/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 70. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(1, 4, No, [1, 3], [2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (1, 4, No, [1, 3], [2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0070/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 71. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(2, 1, No, [0, 2, 4, 5], [1], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (2, 1, No, [0, 2, 4, 5], [1], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0071/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 72. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(3, 4, No, [0, 3], [2, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (3, 4, No, [0, 3], [2, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0072/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 73. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(4, 1, No, [2, 4, 5], [0, 1], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (4, 1, No, [2, 4, 5], [0, 1], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0073/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 74. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 4], [0, 5], [1, 0], [1, 1], [1, 4], [1, 5], [2, 2], [2, 3], [3, 2], [3, 3], [4, 0], [4, 1], [4, 4], [4, 5], [5, 0], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(5, 4, Yes, [0, 1, 4, 5], [0, 1, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (5, 4, Yes, [0, 1, 4, 5], [0, 1, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0074/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 75. Representative Invariance Is Conditional on Relatedness

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [1, 1], [1, 4], [1, 5], [2, 0], [2, 2], [2, 3], [3, 0], [3, 2], [3, 3], [4, 1], [4, 4], [4, 5], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,related?,class_a,class_b,invariance?).

View answer and explanation(0, 1, No, [0, 2, 3], [1, 4, 5], Yes)

Compute both full classes and report the implication aRb implies class_a=class_b. When the representatives are unrelated the implication passes vacuously; that does not assert their two classes are equal. The exact result, retaining the requested order, is (0, 1, No, [0, 2, 3], [1, 4, 5], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0075/)

Equivalence class · qf-mod29-eqclass-representative-invariance · candidate





## 76. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(0, 1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (0, 1, [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0076/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 77. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(1, 4, [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (1, 4, [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0077/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 78. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 3], [2, 4], [2, 5], [3, 0], [3, 2], [3, 3], [3, 4], [3, 5], [4, 0], [4, 2], [4, 3], [4, 4], [4, 5], [5, 0], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(2, 1, [0, 2, 3, 4, 5], [1], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (2, 1, [0, 2, 3, 4, 5], [1], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0078/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 79. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(3, 4, [2, 3, 4, 5], [2, 3, 4, 5], [2, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (3, 4, [2, 3, 4, 5], [2, 3, 4, 5], [2, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0079/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 80. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 2], [3, 3], [3, 4], [3, 5], [4, 2], [4, 3], [4, 4], [4, 5], [5, 2], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(4, 1, [2, 3, 4, 5], [1], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (4, 1, [2, 3, 4, 5], [1], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0080/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 81. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [0, 4], [0, 5], [1, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 0], [3, 1], [3, 3], [3, 4], [3, 5], [4, 0], [4, 1], [4, 3], [4, 4], [4, 5], [5, 0], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(5, 4, [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (5, 4, [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], [0, 1, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0081/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 82. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [1, 3], [1, 4], [1, 5], [2, 0], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(0, 1, [0, 2], [1, 3, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (0, 1, [0, 2], [1, 3, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0082/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 83. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [1, 4], [1, 5], [2, 2], [3, 1], [3, 3], [3, 4], [3, 5], [4, 1], [4, 3], [4, 4], [4, 5], [5, 1], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(1, 4, [1, 3, 4, 5], [1, 3, 4, 5], [1, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (1, 4, [1, 3, 4, 5], [1, 3, 4, 5], [1, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0083/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 84. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [1, 2], [2, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(2, 1, [1, 2], [1, 2], [1, 2], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (2, 1, [1, 2], [1, 2], [1, 2], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0084/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 85. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(3, 4, [3, 4, 5], [3, 4, 5], [3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (3, 4, [3, 4, 5], [3, 4, 5], [3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0085/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 86. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [2, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(4, 1, [3, 4, 5], [1, 2], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (4, 1, [3, 4, 5], [1, 2], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0086/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 87. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [0, 4], [0, 5], [1, 1], [2, 2], [3, 0], [3, 3], [3, 4], [3, 5], [4, 0], [4, 3], [4, 4], [4, 5], [5, 0], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(5, 4, [0, 3, 4, 5], [0, 3, 4, 5], [0, 3, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (5, 4, [0, 3, 4, 5], [0, 3, 4, 5], [0, 3, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0087/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 88. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [1, 1], [2, 0], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(0, 1, [0, 2], [1], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (0, 1, [0, 2], [1], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0088/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 89. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [3, 3], [3, 4], [3, 5], [4, 3], [4, 4], [4, 5], [5, 3], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(1, 4, [0, 1], [3, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (1, 4, [0, 1], [3, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0089/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 90. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 2], [0, 4], [0, 5], [1, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 0], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 1], [4, 2], [4, 4], [4, 5], [5, 0], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(2, 1, [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (2, 1, [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], [0, 1, 2, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0090/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 91. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(3, 4, [0, 3], [1, 2, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (3, 4, [0, 3], [1, 2, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0091/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 92. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 2], [1, 4], [1, 5], [2, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 1], [4, 2], [4, 4], [4, 5], [5, 1], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(4, 1, [1, 2, 4, 5], [1, 2, 4, 5], [1, 2, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (4, 1, [1, 2, 4, 5], [1, 2, 4, 5], [1, 2, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0092/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 93. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [1, 3], [2, 0], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(5, 4, [0, 2, 4, 5], [0, 2, 4, 5], [0, 2, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (5, 4, [0, 2, 4, 5], [0, 2, 4, 5], [0, 2, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0093/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 94. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 3], [1, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 0], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(0, 1, [0, 1, 3], [0, 1, 3], [0, 1, 3], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (0, 1, [0, 1, 3], [0, 1, 3], [0, 1, 3], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0094/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 95. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [1, 1], [1, 3], [2, 2], [2, 4], [2, 5], [3, 1], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=1, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(1, 4, [1, 3], [2, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (1, 4, [1, 3], [2, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0095/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 96. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 4], [0, 5], [1, 1], [2, 0], [2, 2], [2, 4], [2, 5], [3, 3], [4, 0], [4, 2], [4, 4], [4, 5], [5, 0], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=2, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(2, 1, [0, 2, 4, 5], [1], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (2, 1, [0, 2, 4, 5], [1], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0096/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 97. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 3], [1, 1], [2, 2], [2, 4], [2, 5], [3, 0], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=3, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(3, 4, [0, 3], [2, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (3, 4, [0, 3], [2, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0097/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 98. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [1, 0], [1, 1], [2, 2], [2, 4], [2, 5], [3, 3], [4, 2], [4, 4], [4, 5], [5, 2], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=4, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(4, 1, [2, 4, 5], [0, 1], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (4, 1, [2, 4, 5], [0, 1], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0098/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 99. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 1], [0, 4], [0, 5], [1, 0], [1, 1], [1, 4], [1, 5], [2, 2], [2, 3], [3, 2], [3, 3], [4, 0], [4, 1], [4, 4], [4, 5], [5, 0], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=5, b=4. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(5, 4, [0, 1, 4, 5], [0, 1, 4, 5], [0, 1, 4, 5], Yes, equal)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (5, 4, [0, 1, 4, 5], [0, 1, 4, 5], [0, 1, 4, 5], Yes, equal).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0099/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate





## 100. Two Certified Classes Are Equal or Disjoint

Use the complete carrier Fin(6)={0,…,5} and complete relation pairs=[[0, 0], [0, 2], [0, 3], [1, 1], [1, 4], [1, 5], [2, 0], [2, 2], [2, 3], [3, 0], [3, 2], [3, 3], [4, 1], [4, 4], [4, 5], [5, 1], [5, 4], [5, 5]]. Before forming classes, certify reflexivity, symmetry and transitivity on this same carrier; sampled pairs or visual clustering do not suffice. A class contains exactly every related carrier element, listed increasingly. Classes are equal or disjoint; no quotient operation or representative-independent algebra is constructed. a=0, b=1. Give (a,b,class_a,class_b,intersection,equal_or_disjoint?,classification). Classification is equal or disjoint.

View answer and explanation(0, 1, [0, 2, 3], [1, 4, 5], [], Yes, disjoint)

Compare the complete classes and their increasing intersection. A nonempty intersection between unequal classes would contradict the already required equivalence axioms, not define an additional class type. The exact result, retaining the requested order, is (0, 1, [0, 2, 3], [1, 4, 5], [], Yes, disjoint).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261029-r07-equivalence-class-0100/)

Equivalence class · qf-mod29-eqclass-equal-or-disjoint · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261029-r07-equivalence-class-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261029-r07-equivalence-class-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261029-r07-equivalence-class-foundations/validation-report.json)