Partial Correctness, Ranking Obligations, and All-Path Termination
This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
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-20261013-advance-20261019-r05-algorithm-correctness-foundations · 2026-10-13 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
2. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
3. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
4. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
5. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
6. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
7. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
8. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
9. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
10. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
11. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
12. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
13. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
14. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
15. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
16. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
17. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
18. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
19. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
20. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
21. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
22. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
23. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
24. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
25. Inductiveness versus Merely Covering Reachable States
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.
View answer and explanation
([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)
Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-invariant-audit · candidate
26. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
27. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
28. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
29. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
30. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
31. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
32. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
33. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
34. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
35. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
36. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
37. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
38. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
39. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
40. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
41. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
42. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
43. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
44. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
45. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
46. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
47. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
48. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
49. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [5], [], Yes, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
50. Partial Correctness and Its Vacuous Case
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).
View answer and explanation
([5], [], [5], No, No)
Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-partial-audit · candidate
51. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[6, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([6, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([6, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
52. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[7, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([7, 5, 4, 3, 2, 0], [], Yes, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([7, 5, 4, 3, 2, 0], [], Yes, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
53. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[8, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([8, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([8, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
54. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[9, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([9, 5, 4, 3, 2, 0], [], Yes, [4], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([9, 5, 4, 3, 2, 0], [], Yes, [4], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
55. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[10, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([10, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([10, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
56. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[11, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([11, 5, 4, 3, 2, 0], [], Yes, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([11, 5, 4, 3, 2, 0], [], Yes, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
57. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[12, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([12, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([12, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
58. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[13, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([13, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([13, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
59. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[14, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([14, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([14, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
60. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[15, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([15, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([15, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
61. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[16, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([16, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([16, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
62. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[17, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([17, 5, 4, 3, 2, 0], [[2, 1]], No, [4], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([17, 5, 4, 3, 2, 0], [[2, 1]], No, [4], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
63. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[18, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([18, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([18, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
64. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[19, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([19, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([19, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
65. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[20, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([20, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([20, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
66. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[21, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([21, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([21, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
67. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[22, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([22, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([22, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
68. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[23, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([23, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([23, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
69. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[24, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([24, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([24, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
70. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[25, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([25, 5, 4, 3, 2, 0], [[1, 1]], No, [4], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([25, 5, 4, 3, 2, 0], [[1, 1]], No, [4], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
71. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[26, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([26, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([26, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
72. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[27, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([27, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([27, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
73. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[28, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([28, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([28, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
74. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[29, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([29, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([29, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
75. Ranking Decrease, Stuck States, and Actual Termination
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[30, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).
View answer and explanation
([30, 5, 4, 3, 2, 0], [[1, 1], [2, 1]], No, [], No, No)
Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([30, 5, 4, 3, 2, 0], [[1, 1], [2, 1]], No, [], No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-ranking-audit · candidate
76. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], No, Yes, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
77. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], No, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
78. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], No, Yes, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
79. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [4], No, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
80. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], No, Yes, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
81. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], No, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], No, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
82. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], No, Yes, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
83. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [], No, Yes, Yes)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], No, Yes, Yes).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
84. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
85. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
86. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
87. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [4], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
88. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
89. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
90. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
91. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
92. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
93. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
94. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
95. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [4], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
96. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
97. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [3], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
98. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
99. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [5], [], Yes, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
100. Partial Correctness, Termination, and Total Correctness
Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).
View answer and explanation
([5], [], [5], No, No, [], Yes, No, No)
Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No).
Reviewed step-by-step derivation
Algorithm correctness · qf-ext19-correctness-total-separation · candidate
Canonical question JSONL · English JSONL · Validation report