# 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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0001/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0002/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0003/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0004/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0005/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0006/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0007/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0008/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0009/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0010/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0011/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0012/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0013/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0014/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0015/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0016/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0017/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0018/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0019/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0020/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0021/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0022/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0023/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0024/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0025/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0026/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0027/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0028/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0029/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0030/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0031/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0032/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0033/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0034/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0035/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0036/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0037/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0038/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0039/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0040/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0041/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0042/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0043/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0044/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0045/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0046/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0047/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0048/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0049/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0050/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0051/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0052/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0053/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0054/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0055/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0056/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0057/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0058/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0059/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0060/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0061/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0062/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0063/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0064/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0065/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0066/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0067/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0068/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0069/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0070/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0071/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0072/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0073/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0074/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0075/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0076/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0077/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0078/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0079/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0080/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0081/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0082/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0083/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0084/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0085/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0086/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0087/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0088/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0089/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0090/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0091/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0092/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0093/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0094/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0095/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0096/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0097/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0098/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0099/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261019-r05-algorithm-correctness-0100/)

Algorithm correctness · qf-ext19-correctness-total-separation · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261019-r05-algorithm-correctness-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261013-advance-20261019-r05-algorithm-correctness-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261019-r05-algorithm-correctness-foundations/validation-report.json)