Use the complete finite undirected loop-free simple graph on Fin(6)={0,…,5} with edges=[[0, 1], [1, 2], [2, 3], [3, 4]]. Normalize each edge increasingly, then sort edges lexicographically; reversed duplicates and loops are forbidden. Keep isolated vertices in the carrier. No directed-graph conclusion is inferred. The total part assignment=[0, 1, 1, 0, 1, 0] uses labels 0 or 1 for every vertex. Its two inverse-image parts are disjoint and cover the whole carrier. Every edge must cross parts; empty parts are allowed in an edgeless graph. Failure of this assignment alone does not prove the graph nonbipartite. edge_index=2 selects the normalized sorted edge list, not the raw input order. Give (edge_index,edge,left_part,right_part,crosses?,whole_partition_proper?).
G001 — Use the complete finite undirected loop-free simple graph on Fin(6)={0,…,5} with edges=[[0, 1], [1, 2], [2, 3], [3, 4]]. Normalize each edge increasingly, then sort edges lexicographically; reversed duplicates and loops are forbidden. Keep isolated vertices in the carrier. No directed-graph conclusion is inferred. The total part assignment=[0, 1, 1, 0, 1, 0] uses labels 0 or 1 for every vertex. Its two inverse-image parts are disjoint and cover the whole carrier. Every edge must cross parts; empty parts are allowed in an edgeless graph. Failure of this assignment alone does not prove the graph nonbipartite. edge_index=2 selects the normalized sorted edge list, not the raw input order. Give (edge_index,edge,left_part,right_part,crosses?,whole_partition_proper?).
Definition and scope
Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.
Normalize endpoints and then sort edges lexicographically.Reversed duplicates are still duplicates and are rejected,as are loops.The complete carrier includes vertices absent from the edge list.
total original part labels=[0, 1, 1, 0, 1, 0];complete carrier Fin(6)
After
part0=[0, 3, 5];part1=[1, 2, 4];disjoint full cover=Yes
Conditions: none
The two inverse images of0 and1 are disjoint and cover every originally labeled vertex.An empty part is allowed for an edgeless graph;isolates remain included.Edge crossing is still a separate obligation.
Check the endpoints of each actual edge.Reused labels at nonadjacent vertices are not conflicts,and one crossing edge does not certify all other edges.
Check the endpoints of each actual edge.Reused labels at nonadjacent vertices are not conflicts,and one crossing edge does not certify all other edges.
Check the endpoints of each actual edge.Reused labels at nonadjacent vertices are not conflicts,and one crossing edge does not certify all other edges.
Check the endpoints of each actual edge.Reused labels at nonadjacent vertices are not conflicts,and one crossing edge does not certify all other edges.
The original index addresses the normalized sorted edge list,not raw input order.Keep the local crossing flag separate from whole-assignment properness.
S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
Result
(2, [2, 3], 1, 0, Yes, No)
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.