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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Finite two-sided ideal · qf-mod28-ideal-trivial-boundary · candidate
Canonical question JSONL · English JSONL · Validation report