# Additive Subgroups, Separate Left/Right Absorption and Trivial Ideals

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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-20261028-r02-finite-two-sided-ideal-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. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 1, 2, 4, 6], 0, No, [[3, 1, 3], [5, 1, 5], [7, 1, 7]], [[1, 3, 3], [1, 5, 5], [1, 7, 7]], No)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 4, 6], 0, No, [[3, 1, 3], [5, 1, 5], [7, 1, 7]], [[1, 3, 3], [1, 5, 5], [1, 7, 7]], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0001/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 2. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0002/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 3. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0003/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 4. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 2, 4, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0004/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 5. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0005/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 6. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 1, 2, 4, 6], 0, No, [[3, 1, 3], [5, 1, 5], [7, 1, 7]], [[1, 3, 3], [1, 5, 5], [1, 7, 7]], No)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 4, 6], 0, No, [[3, 1, 3], [5, 1, 5], [7, 1, 7]], [[1, 3, 3], [1, 5, 5], [1, 7, 7]], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0006/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 7. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0007/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 8. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0008/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 9. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 5], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0009/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 10. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0010/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 11. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 1, 2, 4, 5], 0, No, [[3, 1, 3], [6, 1, 6], [7, 1, 7]], [[1, 3, 3], [1, 6, 6], [1, 7, 7]], No)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 4, 5], 0, No, [[3, 1, 3], [6, 1, 6], [7, 1, 7]], [[1, 3, 3], [1, 6, 6], [1, 7, 7]], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0011/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 12. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0012/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 13. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0013/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 14. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 2, 6, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 6, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0014/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 15. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 6], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0015/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 16. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 1, 2, 6, 7], 0, No, [[3, 1, 3], [4, 1, 4], [5, 1, 5]], [[1, 3, 3], [1, 4, 4], [1, 5, 5]], No)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 6, 7], 0, No, [[3, 1, 3], [4, 1, 4], [5, 1, 5]], [[1, 3, 3], [1, 4, 4], [1, 5, 5]], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0016/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 17. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 6], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0017/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 18. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0018/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 19. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 2, 5, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0019/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 20. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0020/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 21. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 1, 2, 4, 7], 0, No, [[3, 1, 3], [5, 1, 5], [6, 1, 6]], [[1, 3, 3], [1, 5, 5], [1, 6, 6]], No)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 4, 7], 0, No, [[3, 1, 3], [5, 1, 5], [6, 1, 6]], [[1, 3, 3], [1, 5, 5], [1, 6, 6]], No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0021/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 22. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0022/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 23. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0023/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 24. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

View answer and explanation([0, 2, 5, 7], 0, Yes, [], [], Yes)

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0024/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 25. Two-Sided Ideals Require Three Complete Qualifications

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 3] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,additive_subgroup?,missing_left,missing_right,two_sided_ideal?). Missing-left rows are [r,x,r*x]; missing-right rows are [x,r,x*r], ordered by parent label r then subset label x.

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

Check nonempty additive-subgroup closure, including zero and inverses, before conjoining both absorption audits over every parent element. A one-sided ideal or subring is not automatically two-sided. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 3], 0, Yes, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0025/)

Finite two-sided ideal · qf-mod28-ideal-qualification · candidate





## 26. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 4, 6], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 6], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0026/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 27. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0027/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 28. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0028/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 29. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 4, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0029/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 30. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0030/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 31. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 4, 6], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 6], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0031/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 32. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0032/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 33. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0033/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 34. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 5], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0034/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 35. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0035/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 36. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 5], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0036/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 37. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0037/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 38. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0038/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 39. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 6, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 6, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0039/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 40. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 6], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0040/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 41. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 6, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 6, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0041/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 42. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 6], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0042/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 43. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0043/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 44. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 5, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0044/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 45. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0045/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 46. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 4, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0046/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 47. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0047/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 48. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0048/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 49. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

View answer and explanation([0, 2, 5, 7], 0, [], [], Yes)

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0049/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 50. Check the Additive Structure before Multiplicative Absorption

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 3] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero,all missing [a,b,a+b] rows,all missing [a,additive_inverse] rows,additive_subgroup?).

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

Use the normalized subset order, testing every ordered addition pair and each additive inverse. The subset must also be nonempty and contain the actual additive zero. No absorption or ideal claim follows from this flag alone. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 3], 0, [], [], Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0050/)

Finite two-sided ideal · qf-mod28-ideal-additive-subgroup · candidate





## 51. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 4, 6], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 6], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0051/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 52. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0052/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 53. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0053/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 54. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 4, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0054/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 55. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 4] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 4], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0055/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 56. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 4, 6], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 6], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0056/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 57. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0057/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 58. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0058/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 59. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 5], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0059/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 60. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0060/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 61. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 5], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0061/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 62. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 5] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 5], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0062/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 63. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0063/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 64. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 6, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 6, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0064/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 65. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 6], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0065/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 66. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 6, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 6, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0066/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 67. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 6] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 6], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 6], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0067/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 68. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0068/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 69. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 5, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0069/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 70. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0070/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 71. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 4, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 4, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 4, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0071/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 72. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0072/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 73. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0073/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 74. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 2, 5, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

View answer and explanation([0, 2, 5, 7], [], [], Yes, Yes)

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 2, 5, 7], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0074/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 75. Left Absorption and Right Absorption Must Both Be Reported

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 3] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,missing_left [r,x,r*x] rows,missing_right [x,r,x*r] rows,left_absorbing?,right_absorbing?). Both failure lists use parent-r outer order and subset-x inner order.

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

Test each product on both sides using all parent-ring elements, even in a noncommutative ring. An empty subset passes these universal absorption checks vacuously but fails additive-subgroup and ideal qualification. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 3], [], [], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0075/)

Finite two-sided ideal · qf-mod28-ideal-two-sided-absorption · candidate





## 76. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0076/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 77. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0077/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 78. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0078/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 79. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0079/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 80. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0080/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 81. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0081/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 82. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0082/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 83. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0083/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 84. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0084/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 85. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0085/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 86. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0086/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 87. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0087/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 88. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0088/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 89. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0089/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 90. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0090/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 91. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0091/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 92. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0092/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 93. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0093/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 94. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0094/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 95. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0095/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 96. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0096/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 97. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0097/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 98. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0098/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 99. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0, 1, 2, 3, 4, 5, 6, 7] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals)

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], No, Yes, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0099/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate





## 100. The Zero Ideal and Whole Ring Are Not the Empty Subset

Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Normalize subset=[0] as distinct carrier labels in increasing order. Its operations and inverse map come from this same parent ring. Additive-subgroup qualification, left absorption and right absorption are independent obligations. No quotient, maximality or primality is constructed. Give (subset,zero_ideal?,whole_ring?,qualified?,zero_and_whole_are_always_two_sided_ideals).

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

Compare with the singleton containing the actual additive zero and with the full carrier separately. These two subsets are always two-sided ideals; the supplied subset is still audited rather than assumed to be one of them. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is ([0], Yes, No, Yes, zero_and_whole_are_always_two_sided_ideals).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261028-r02-finite-two-sided-ideal-0100/)

Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261028-r02-finite-two-sided-ideal-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261028-r02-finite-two-sided-ideal-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261028-r02-finite-two-sided-ideal-foundations/validation-report.json)