# Exact Tape Traces, Canonical Global Rankings and Explicit Recognizer Loops

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-20261015-advance-20261113-r05-turing-machine-decision-model-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q00, q10, qa0, qr0], input_alphabet=[00], tape_alphabet=[00, X0, _0], blank=_0, start=q00, accept=qa0, reject=qr0, transitions=[[q00, 00, q10, X0, R], [q00, _0, qr0, _0, S], [q10, _0, qa0, _0, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q00, q10, qa0, qr0], [00], [00, X0, _0], _0, q00, qa0, qr0, [[q00, _0, qr0, _0, S], [q00, 00, q10, X0, R], [q10, _0, qa0, _0, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q00, q10, qa0, qr0], [00], [00, X0, _0], _0, q00, qa0, qr0, [[q00, _0, qr0, _0, S], [q00, 00, q10, X0, R], [q10, _0, qa0, _0, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0001/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 2. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q01, q11, qa1, qr1], input_alphabet=[01], tape_alphabet=[01, X1, _1], blank=_1, start=q01, accept=qa1, reject=qr1, transitions=[[q01, 01, q11, X1, R], [q01, _1, qr1, _1, S], [q11, _1, qa1, _1, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q01, q11, qa1, qr1], [01], [01, X1, _1], _1, q01, qa1, qr1, [[q01, _1, qr1, _1, S], [q01, 01, q11, X1, R], [q11, _1, qa1, _1, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q01, q11, qa1, qr1], [01], [01, X1, _1], _1, q01, qa1, qr1, [[q01, _1, qr1, _1, S], [q01, 01, q11, X1, R], [q11, _1, qa1, _1, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0002/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 3. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q02, q12, qa2, qr2], input_alphabet=[02], tape_alphabet=[02, X2, _2], blank=_2, start=q02, accept=qa2, reject=qr2, transitions=[[q02, 02, q12, X2, R], [q02, _2, qr2, _2, S], [q12, _2, qa2, _2, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q02, q12, qa2, qr2], [02], [02, X2, _2], _2, q02, qa2, qr2, [[q02, _2, qr2, _2, S], [q02, 02, q12, X2, R], [q12, _2, qa2, _2, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q02, q12, qa2, qr2], [02], [02, X2, _2], _2, q02, qa2, qr2, [[q02, _2, qr2, _2, S], [q02, 02, q12, X2, R], [q12, _2, qa2, _2, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0003/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 4. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q03, q13, qa3, qr3], input_alphabet=[03], tape_alphabet=[03, X3, _3], blank=_3, start=q03, accept=qa3, reject=qr3, transitions=[[q03, 03, q13, X3, R], [q03, _3, qr3, _3, S], [q13, _3, qa3, _3, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q03, q13, qa3, qr3], [03], [03, X3, _3], _3, q03, qa3, qr3, [[q03, _3, qr3, _3, S], [q03, 03, q13, X3, R], [q13, _3, qa3, _3, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q03, q13, qa3, qr3], [03], [03, X3, _3], _3, q03, qa3, qr3, [[q03, _3, qr3, _3, S], [q03, 03, q13, X3, R], [q13, _3, qa3, _3, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0004/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 5. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q04, q14, qa4, qr4], input_alphabet=[04], tape_alphabet=[04, X4, _4], blank=_4, start=q04, accept=qa4, reject=qr4, transitions=[[q04, 04, q14, X4, R], [q04, _4, qr4, _4, S], [q14, _4, qa4, _4, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q04, q14, qa4, qr4], [04], [04, X4, _4], _4, q04, qa4, qr4, [[q04, _4, qr4, _4, S], [q04, 04, q14, X4, R], [q14, _4, qa4, _4, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q04, q14, qa4, qr4], [04], [04, X4, _4], _4, q04, qa4, qr4, [[q04, _4, qr4, _4, S], [q04, 04, q14, X4, R], [q14, _4, qa4, _4, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0005/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 6. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q05, q15, qa5, qr5], input_alphabet=[05], tape_alphabet=[05, X5, _5], blank=_5, start=q05, accept=qa5, reject=qr5, transitions=[[q05, 05, q15, X5, R], [q05, _5, qr5, _5, S], [q15, _5, qa5, _5, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q05, q15, qa5, qr5], [05], [05, X5, _5], _5, q05, qa5, qr5, [[q05, _5, qr5, _5, S], [q05, 05, q15, X5, R], [q15, _5, qa5, _5, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q05, q15, qa5, qr5], [05], [05, X5, _5], _5, q05, qa5, qr5, [[q05, _5, qr5, _5, S], [q05, 05, q15, X5, R], [q15, _5, qa5, _5, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0006/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 7. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q06, q16, qa6, qr6], input_alphabet=[06], tape_alphabet=[06, X6, _6], blank=_6, start=q06, accept=qa6, reject=qr6, transitions=[[q06, 06, q16, X6, R], [q06, _6, qr6, _6, S], [q16, _6, qa6, _6, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q06, q16, qa6, qr6], [06], [06, X6, _6], _6, q06, qa6, qr6, [[q06, _6, qr6, _6, S], [q06, 06, q16, X6, R], [q16, _6, qa6, _6, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q06, q16, qa6, qr6], [06], [06, X6, _6], _6, q06, qa6, qr6, [[q06, _6, qr6, _6, S], [q06, 06, q16, X6, R], [q16, _6, qa6, _6, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0007/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 8. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q07, q17, qa7, qr7], input_alphabet=[07], tape_alphabet=[07, X7, _7], blank=_7, start=q07, accept=qa7, reject=qr7, transitions=[[q07, 07, q17, X7, R], [q07, _7, qr7, _7, S], [q17, _7, qa7, _7, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q07, q17, qa7, qr7], [07], [07, X7, _7], _7, q07, qa7, qr7, [[q07, _7, qr7, _7, S], [q07, 07, q17, X7, R], [q17, _7, qa7, _7, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q07, q17, qa7, qr7], [07], [07, X7, _7], _7, q07, qa7, qr7, [[q07, _7, qr7, _7, S], [q07, 07, q17, X7, R], [q17, _7, qa7, _7, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0008/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 9. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q08, q18, qa8, qr8], input_alphabet=[08], tape_alphabet=[08, X8, _8], blank=_8, start=q08, accept=qa8, reject=qr8, transitions=[[q08, 08, q18, X8, R], [q08, _8, qr8, _8, S], [q18, _8, qa8, _8, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q08, q18, qa8, qr8], [08], [08, X8, _8], _8, q08, qa8, qr8, [[q08, _8, qr8, _8, S], [q08, 08, q18, X8, R], [q18, _8, qa8, _8, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q08, q18, qa8, qr8], [08], [08, X8, _8], _8, q08, qa8, qr8, [[q08, _8, qr8, _8, S], [q08, 08, q18, X8, R], [q18, _8, qa8, _8, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0009/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 10. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q09, q19, qa9, qr9], input_alphabet=[09], tape_alphabet=[09, X9, _9], blank=_9, start=q09, accept=qa9, reject=qr9, transitions=[[q09, 09, q19, X9, R], [q09, _9, qr9, _9, S], [q19, _9, qa9, _9, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q09, q19, qa9, qr9], [09], [09, X9, _9], _9, q09, qa9, qr9, [[q09, _9, qr9, _9, S], [q09, 09, q19, X9, R], [q19, _9, qa9, _9, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q09, q19, qa9, qr9], [09], [09, X9, _9], _9, q09, qa9, qr9, [[q09, _9, qr9, _9, S], [q09, 09, q19, X9, R], [q19, _9, qa9, _9, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0010/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 11. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q010, q110, qa10, qr10], input_alphabet=[010], tape_alphabet=[010, X10, _10], blank=_10, start=q010, accept=qa10, reject=qr10, transitions=[[q010, 010, q110, X10, R], [q010, _10, qr10, _10, S], [q110, _10, qa10, _10, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q010, q110, qa10, qr10], [010], [010, X10, _10], _10, q010, qa10, qr10, [[q010, _10, qr10, _10, S], [q010, 010, q110, X10, R], [q110, _10, qa10, _10, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q010, q110, qa10, qr10], [010], [010, X10, _10], _10, q010, qa10, qr10, [[q010, _10, qr10, _10, S], [q010, 010, q110, X10, R], [q110, _10, qa10, _10, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0011/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 12. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q011, q111, qa11, qr11], input_alphabet=[011], tape_alphabet=[011, X11, _11], blank=_11, start=q011, accept=qa11, reject=qr11, transitions=[[q011, 011, q111, X11, R], [q011, _11, qr11, _11, S], [q111, _11, qa11, _11, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q011, q111, qa11, qr11], [011], [011, X11, _11], _11, q011, qa11, qr11, [[q011, _11, qr11, _11, S], [q011, 011, q111, X11, R], [q111, _11, qa11, _11, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q011, q111, qa11, qr11], [011], [011, X11, _11], _11, q011, qa11, qr11, [[q011, _11, qr11, _11, S], [q011, 011, q111, X11, R], [q111, _11, qa11, _11, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0012/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 13. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q012, q112, qa12, qr12], input_alphabet=[012], tape_alphabet=[012, X12, _12], blank=_12, start=q012, accept=qa12, reject=qr12, transitions=[[q012, 012, q112, X12, R], [q012, _12, qr12, _12, S], [q112, _12, qa12, _12, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q012, q112, qa12, qr12], [012], [012, X12, _12], _12, q012, qa12, qr12, [[q012, _12, qr12, _12, S], [q012, 012, q112, X12, R], [q112, _12, qa12, _12, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q012, q112, qa12, qr12], [012], [012, X12, _12], _12, q012, qa12, qr12, [[q012, _12, qr12, _12, S], [q012, 012, q112, X12, R], [q112, _12, qa12, _12, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0013/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 14. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q013, q113, qa13, qr13], input_alphabet=[013], tape_alphabet=[013, X13, _13], blank=_13, start=q013, accept=qa13, reject=qr13, transitions=[[q013, 013, q113, X13, R], [q013, _13, qr13, _13, S], [q113, _13, qa13, _13, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q013, q113, qa13, qr13], [013], [013, X13, _13], _13, q013, qa13, qr13, [[q013, _13, qr13, _13, S], [q013, 013, q113, X13, R], [q113, _13, qa13, _13, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q013, q113, qa13, qr13], [013], [013, X13, _13], _13, q013, qa13, qr13, [[q013, _13, qr13, _13, S], [q013, 013, q113, X13, R], [q113, _13, qa13, _13, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0014/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 15. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q014, q114, qa14, qr14], input_alphabet=[014], tape_alphabet=[014, X14, _14], blank=_14, start=q014, accept=qa14, reject=qr14, transitions=[[q014, 014, q114, X14, R], [q014, _14, qr14, _14, S], [q114, _14, qa14, _14, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q014, q114, qa14, qr14], [014], [014, X14, _14], _14, q014, qa14, qr14, [[q014, _14, qr14, _14, S], [q014, 014, q114, X14, R], [q114, _14, qa14, _14, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q014, q114, qa14, qr14], [014], [014, X14, _14], _14, q014, qa14, qr14, [[q014, _14, qr14, _14, S], [q014, 014, q114, X14, R], [q114, _14, qa14, _14, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0015/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 16. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q015, q115, qa15, qr15], input_alphabet=[015], tape_alphabet=[015, X15, _15], blank=_15, start=q015, accept=qa15, reject=qr15, transitions=[[q015, 015, q115, X15, R], [q015, _15, qr15, _15, S], [q115, _15, qa15, _15, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q015, q115, qa15, qr15], [015], [015, X15, _15], _15, q015, qa15, qr15, [[q015, _15, qr15, _15, S], [q015, 015, q115, X15, R], [q115, _15, qa15, _15, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q015, q115, qa15, qr15], [015], [015, X15, _15], _15, q015, qa15, qr15, [[q015, _15, qr15, _15, S], [q015, 015, q115, X15, R], [q115, _15, qa15, _15, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0016/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 17. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q016, q116, qa16, qr16], input_alphabet=[016], tape_alphabet=[016, X16, _16], blank=_16, start=q016, accept=qa16, reject=qr16, transitions=[[q016, 016, q116, X16, R], [q016, _16, qr16, _16, S], [q116, _16, qa16, _16, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q016, q116, qa16, qr16], [016], [016, X16, _16], _16, q016, qa16, qr16, [[q016, _16, qr16, _16, S], [q016, 016, q116, X16, R], [q116, _16, qa16, _16, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q016, q116, qa16, qr16], [016], [016, X16, _16], _16, q016, qa16, qr16, [[q016, _16, qr16, _16, S], [q016, 016, q116, X16, R], [q116, _16, qa16, _16, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0017/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 18. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q017, q117, qa17, qr17], input_alphabet=[017], tape_alphabet=[017, X17, _17], blank=_17, start=q017, accept=qa17, reject=qr17, transitions=[[q017, 017, q117, X17, R], [q017, _17, qr17, _17, S], [q117, _17, qa17, _17, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q017, q117, qa17, qr17], [017], [017, X17, _17], _17, q017, qa17, qr17, [[q017, _17, qr17, _17, S], [q017, 017, q117, X17, R], [q117, _17, qa17, _17, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q017, q117, qa17, qr17], [017], [017, X17, _17], _17, q017, qa17, qr17, [[q017, _17, qr17, _17, S], [q017, 017, q117, X17, R], [q117, _17, qa17, _17, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0018/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 19. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q018, q118, qa18, qr18], input_alphabet=[018], tape_alphabet=[018, X18, _18], blank=_18, start=q018, accept=qa18, reject=qr18, transitions=[[q018, 018, q118, X18, R], [q018, _18, qr18, _18, S], [q118, _18, qa18, _18, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q018, q118, qa18, qr18], [018], [018, X18, _18], _18, q018, qa18, qr18, [[q018, _18, qr18, _18, S], [q018, 018, q118, X18, R], [q118, _18, qa18, _18, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q018, q118, qa18, qr18], [018], [018, X18, _18], _18, q018, qa18, qr18, [[q018, _18, qr18, _18, S], [q018, 018, q118, X18, R], [q118, _18, qa18, _18, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0019/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 20. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q019, q119, qa19, qr19], input_alphabet=[019], tape_alphabet=[019, X19, _19], blank=_19, start=q019, accept=qa19, reject=qr19, transitions=[[q019, 019, q119, X19, R], [q019, _19, qr19, _19, S], [q119, _19, qa19, _19, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q019, q119, qa19, qr19], [019], [019, X19, _19], _19, q019, qa19, qr19, [[q019, _19, qr19, _19, S], [q019, 019, q119, X19, R], [q119, _19, qa19, _19, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q019, q119, qa19, qr19], [019], [019, X19, _19], _19, q019, qa19, qr19, [[q019, _19, qr19, _19, S], [q019, 019, q119, X19, R], [q119, _19, qa19, _19, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0020/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 21. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q020, q120, qa20, qr20], input_alphabet=[020], tape_alphabet=[020, X20, _20], blank=_20, start=q020, accept=qa20, reject=qr20, transitions=[[q020, 020, q120, X20, R], [q020, _20, qr20, _20, S], [q120, _20, qa20, _20, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q020, q120, qa20, qr20], [020], [020, X20, _20], _20, q020, qa20, qr20, [[q020, _20, qr20, _20, S], [q020, 020, q120, X20, R], [q120, _20, qa20, _20, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q020, q120, qa20, qr20], [020], [020, X20, _20], _20, q020, qa20, qr20, [[q020, _20, qr20, _20, S], [q020, 020, q120, X20, R], [q120, _20, qa20, _20, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0021/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 22. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q021, q121, qa21, qr21], input_alphabet=[021], tape_alphabet=[021, X21, _21], blank=_21, start=q021, accept=qa21, reject=qr21, transitions=[[q021, 021, q121, X21, R], [q021, _21, qr21, _21, S], [q121, _21, qa21, _21, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q021, q121, qa21, qr21], [021], [021, X21, _21], _21, q021, qa21, qr21, [[q021, _21, qr21, _21, S], [q021, 021, q121, X21, R], [q121, _21, qa21, _21, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q021, q121, qa21, qr21], [021], [021, X21, _21], _21, q021, qa21, qr21, [[q021, _21, qr21, _21, S], [q021, 021, q121, X21, R], [q121, _21, qa21, _21, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0022/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 23. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q022, q122, qa22, qr22], input_alphabet=[022], tape_alphabet=[022, X22, _22], blank=_22, start=q022, accept=qa22, reject=qr22, transitions=[[q022, 022, q122, X22, R], [q022, _22, qr22, _22, S], [q122, _22, qa22, _22, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q022, q122, qa22, qr22], [022], [022, X22, _22], _22, q022, qa22, qr22, [[q022, _22, qr22, _22, S], [q022, 022, q122, X22, R], [q122, _22, qa22, _22, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q022, q122, qa22, qr22], [022], [022, X22, _22], _22, q022, qa22, qr22, [[q022, _22, qr22, _22, S], [q022, 022, q122, X22, R], [q122, _22, qa22, _22, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0023/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 24. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q023, q123, qa23, qr23], input_alphabet=[023], tape_alphabet=[023, X23, _23], blank=_23, start=q023, accept=qa23, reject=qr23, transitions=[[q023, 023, q123, X23, R], [q023, _23, qr23, _23, S], [q123, _23, qa23, _23, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q023, q123, qa23, qr23], [023], [023, X23, _23], _23, q023, qa23, qr23, [[q023, _23, qr23, _23, S], [q023, 023, q123, X23, R], [q123, _23, qa23, _23, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q023, q123, qa23, qr23], [023], [023, X23, _23], _23, q023, qa23, qr23, [[q023, _23, qr23, _23, S], [q023, 023, q123, X23, R], [q123, _23, qa23, _23, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0024/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 25. A Qualified Deterministic Single-Tape Signature Is Not Yet a Global Decider

Use deterministic single-tape states=[q024, q124, qa24, qr24], input_alphabet=[024], tape_alphabet=[024, X24, _24], blank=_24, start=q024, accept=qa24, reject=qr24, transitions=[[q024, 024, q124, X24, R], [q024, _24, qr24, _24, S], [q124, _24, qa24, _24, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, qualified, convention).

View answer and explanation([q024, q124, qa24, qr24], [024], [024, X24, _24], _24, q024, qa24, qr24, [[q024, _24, qr24, _24, S], [q024, 024, q124, X24, R], [q124, _24, qa24, _24, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions)

Check finite alphabets/states, blank/start/halting states, deterministic keys and every transition field. A partial transition table is permitted at signature level; absence of a rule is not silently filled with rejection. Execution and all-input total correctness remain separate. The exact result, retaining the requested order, is ([q024, q124, qa24, qr24], [024], [024, X24, _24], _24, q024, qa24, qr24, [[q024, _24, qr24, _24, S], [q024, 024, q124, X24, R], [q124, _24, qa24, _24, L]], Yes, deterministic_single_tape_signature_with_halting_states_without_outgoing_transitions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0025/)

Turing-machine decision model · qf-mod44-tm-signature-qualification · candidate





## 26. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q00, q10, qa0, qr0], input_alphabet=[00], tape_alphabet=[00, X0, _0], blank=_0, start=q00, accept=qa0, reject=qr0, transitions=[[q00, 00, q10, X0, R], [q00, _0, qr0, _0, S], [q10, _0, qa0, _0, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[00], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([00], 2, [[q00, 0, [[0, 00]]], [q10, 1, [[0, X0]]], [qa0, 0, [[0, X0]]]], [step, step], qa0, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([00], 2, [[q00, 0, [[0, 00]]], [q10, 1, [[0, X0]]], [qa0, 0, [[0, X0]]]], [step, step], qa0, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0026/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 27. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q01, q11, qa1, qr1], input_alphabet=[01], tape_alphabet=[01, X1, _1], blank=_1, start=q01, accept=qa1, reject=qr1, transitions=[[q01, 01, q11, X1, R], [q01, _1, qr1, _1, S], [q11, _1, qa1, _1, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[01], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([01], 2, [[q01, 0, [[0, 01]]], [q11, 1, [[0, X1]]], [qa1, 0, [[0, X1]]]], [step, step], qa1, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([01], 2, [[q01, 0, [[0, 01]]], [q11, 1, [[0, X1]]], [qa1, 0, [[0, X1]]]], [step, step], qa1, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0027/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 28. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q02, q12, qa2, qr2], input_alphabet=[02], tape_alphabet=[02, X2, _2], blank=_2, start=q02, accept=qa2, reject=qr2, transitions=[[q02, 02, q12, X2, R], [q02, _2, qr2, _2, S], [q12, _2, qa2, _2, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[02], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([02], 2, [[q02, 0, [[0, 02]]], [q12, 1, [[0, X2]]], [qa2, 0, [[0, X2]]]], [step, step], qa2, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([02], 2, [[q02, 0, [[0, 02]]], [q12, 1, [[0, X2]]], [qa2, 0, [[0, X2]]]], [step, step], qa2, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0028/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 29. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q03, q13, qa3, qr3], input_alphabet=[03], tape_alphabet=[03, X3, _3], blank=_3, start=q03, accept=qa3, reject=qr3, transitions=[[q03, 03, q13, X3, R], [q03, _3, qr3, _3, S], [q13, _3, qa3, _3, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[03], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([03], 2, [[q03, 0, [[0, 03]]], [q13, 1, [[0, X3]]], [qa3, 0, [[0, X3]]]], [step, step], qa3, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([03], 2, [[q03, 0, [[0, 03]]], [q13, 1, [[0, X3]]], [qa3, 0, [[0, X3]]]], [step, step], qa3, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0029/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 30. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q04, q14, qa4, qr4], input_alphabet=[04], tape_alphabet=[04, X4, _4], blank=_4, start=q04, accept=qa4, reject=qr4, transitions=[[q04, 04, q14, X4, R], [q04, _4, qr4, _4, S], [q14, _4, qa4, _4, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[04], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([04], 2, [[q04, 0, [[0, 04]]], [q14, 1, [[0, X4]]], [qa4, 0, [[0, X4]]]], [step, step], qa4, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([04], 2, [[q04, 0, [[0, 04]]], [q14, 1, [[0, X4]]], [qa4, 0, [[0, X4]]]], [step, step], qa4, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0030/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 31. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q05, q15, qa5, qr5], input_alphabet=[05], tape_alphabet=[05, X5, _5], blank=_5, start=q05, accept=qa5, reject=qr5, transitions=[[q05, 05, q15, X5, R], [q05, _5, qr5, _5, S], [q15, _5, qa5, _5, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[05], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([05], 2, [[q05, 0, [[0, 05]]], [q15, 1, [[0, X5]]], [qa5, 0, [[0, X5]]]], [step, step], qa5, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([05], 2, [[q05, 0, [[0, 05]]], [q15, 1, [[0, X5]]], [qa5, 0, [[0, X5]]]], [step, step], qa5, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0031/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 32. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q06, q16, qa6, qr6], input_alphabet=[06], tape_alphabet=[06, X6, _6], blank=_6, start=q06, accept=qa6, reject=qr6, transitions=[[q06, 06, q16, X6, R], [q06, _6, qr6, _6, S], [q16, _6, qa6, _6, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[06], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([06], 2, [[q06, 0, [[0, 06]]], [q16, 1, [[0, X6]]], [qa6, 0, [[0, X6]]]], [step, step], qa6, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([06], 2, [[q06, 0, [[0, 06]]], [q16, 1, [[0, X6]]], [qa6, 0, [[0, X6]]]], [step, step], qa6, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0032/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 33. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q07, q17, qa7, qr7], input_alphabet=[07], tape_alphabet=[07, X7, _7], blank=_7, start=q07, accept=qa7, reject=qr7, transitions=[[q07, 07, q17, X7, R], [q07, _7, qr7, _7, S], [q17, _7, qa7, _7, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[07], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([07], 2, [[q07, 0, [[0, 07]]], [q17, 1, [[0, X7]]], [qa7, 0, [[0, X7]]]], [step, step], qa7, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([07], 2, [[q07, 0, [[0, 07]]], [q17, 1, [[0, X7]]], [qa7, 0, [[0, X7]]]], [step, step], qa7, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0033/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 34. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q08, q18, qa8, qr8], input_alphabet=[08], tape_alphabet=[08, X8, _8], blank=_8, start=q08, accept=qa8, reject=qr8, transitions=[[q08, 08, q18, X8, R], [q08, _8, qr8, _8, S], [q18, _8, qa8, _8, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[08], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([08], 2, [[q08, 0, [[0, 08]]], [q18, 1, [[0, X8]]], [qa8, 0, [[0, X8]]]], [step, step], qa8, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([08], 2, [[q08, 0, [[0, 08]]], [q18, 1, [[0, X8]]], [qa8, 0, [[0, X8]]]], [step, step], qa8, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0034/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 35. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q09, q19, qa9, qr9], input_alphabet=[09], tape_alphabet=[09, X9, _9], blank=_9, start=q09, accept=qa9, reject=qr9, transitions=[[q09, 09, q19, X9, R], [q09, _9, qr9, _9, S], [q19, _9, qa9, _9, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[09], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([09], 2, [[q09, 0, [[0, 09]]], [q19, 1, [[0, X9]]], [qa9, 0, [[0, X9]]]], [step, step], qa9, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([09], 2, [[q09, 0, [[0, 09]]], [q19, 1, [[0, X9]]], [qa9, 0, [[0, X9]]]], [step, step], qa9, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0035/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 36. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q010, q110, qa10, qr10], input_alphabet=[010], tape_alphabet=[010, X10, _10], blank=_10, start=q010, accept=qa10, reject=qr10, transitions=[[q010, 010, q110, X10, R], [q010, _10, qr10, _10, S], [q110, _10, qa10, _10, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[010], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([010], 2, [[q010, 0, [[0, 010]]], [q110, 1, [[0, X10]]], [qa10, 0, [[0, X10]]]], [step, step], qa10, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([010], 2, [[q010, 0, [[0, 010]]], [q110, 1, [[0, X10]]], [qa10, 0, [[0, X10]]]], [step, step], qa10, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0036/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 37. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q011, q111, qa11, qr11], input_alphabet=[011], tape_alphabet=[011, X11, _11], blank=_11, start=q011, accept=qa11, reject=qr11, transitions=[[q011, 011, q111, X11, R], [q011, _11, qr11, _11, S], [q111, _11, qa11, _11, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[011], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([011], 2, [[q011, 0, [[0, 011]]], [q111, 1, [[0, X11]]], [qa11, 0, [[0, X11]]]], [step, step], qa11, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([011], 2, [[q011, 0, [[0, 011]]], [q111, 1, [[0, X11]]], [qa11, 0, [[0, X11]]]], [step, step], qa11, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0037/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 38. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q012, q112, qa12, qr12], input_alphabet=[012], tape_alphabet=[012, X12, _12], blank=_12, start=q012, accept=qa12, reject=qr12, transitions=[[q012, 012, q112, X12, R], [q012, _12, qr12, _12, S], [q112, _12, qa12, _12, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[012], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([012], 2, [[q012, 0, [[0, 012]]], [q112, 1, [[0, X12]]], [qa12, 0, [[0, X12]]]], [step, step], qa12, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([012], 2, [[q012, 0, [[0, 012]]], [q112, 1, [[0, X12]]], [qa12, 0, [[0, X12]]]], [step, step], qa12, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0038/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 39. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q013, q113, qa13, qr13], input_alphabet=[013], tape_alphabet=[013, X13, _13], blank=_13, start=q013, accept=qa13, reject=qr13, transitions=[[q013, 013, q113, X13, R], [q013, _13, qr13, _13, S], [q113, _13, qa13, _13, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[013], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([013], 2, [[q013, 0, [[0, 013]]], [q113, 1, [[0, X13]]], [qa13, 0, [[0, X13]]]], [step, step], qa13, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([013], 2, [[q013, 0, [[0, 013]]], [q113, 1, [[0, X13]]], [qa13, 0, [[0, X13]]]], [step, step], qa13, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0039/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 40. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q014, q114, qa14, qr14], input_alphabet=[014], tape_alphabet=[014, X14, _14], blank=_14, start=q014, accept=qa14, reject=qr14, transitions=[[q014, 014, q114, X14, R], [q014, _14, qr14, _14, S], [q114, _14, qa14, _14, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[014], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([014], 2, [[q014, 0, [[0, 014]]], [q114, 1, [[0, X14]]], [qa14, 0, [[0, X14]]]], [step, step], qa14, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([014], 2, [[q014, 0, [[0, 014]]], [q114, 1, [[0, X14]]], [qa14, 0, [[0, X14]]]], [step, step], qa14, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0040/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 41. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q015, q115, qa15, qr15], input_alphabet=[015], tape_alphabet=[015, X15, _15], blank=_15, start=q015, accept=qa15, reject=qr15, transitions=[[q015, 015, q115, X15, R], [q015, _15, qr15, _15, S], [q115, _15, qa15, _15, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[015], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([015], 2, [[q015, 0, [[0, 015]]], [q115, 1, [[0, X15]]], [qa15, 0, [[0, X15]]]], [step, step], qa15, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([015], 2, [[q015, 0, [[0, 015]]], [q115, 1, [[0, X15]]], [qa15, 0, [[0, X15]]]], [step, step], qa15, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0041/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 42. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q016, q116, qa16, qr16], input_alphabet=[016], tape_alphabet=[016, X16, _16], blank=_16, start=q016, accept=qa16, reject=qr16, transitions=[[q016, 016, q116, X16, R], [q016, _16, qr16, _16, S], [q116, _16, qa16, _16, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[016], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([016], 2, [[q016, 0, [[0, 016]]], [q116, 1, [[0, X16]]], [qa16, 0, [[0, X16]]]], [step, step], qa16, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([016], 2, [[q016, 0, [[0, 016]]], [q116, 1, [[0, X16]]], [qa16, 0, [[0, X16]]]], [step, step], qa16, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0042/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 43. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q017, q117, qa17, qr17], input_alphabet=[017], tape_alphabet=[017, X17, _17], blank=_17, start=q017, accept=qa17, reject=qr17, transitions=[[q017, 017, q117, X17, R], [q017, _17, qr17, _17, S], [q117, _17, qa17, _17, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[017], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([017], 2, [[q017, 0, [[0, 017]]], [q117, 1, [[0, X17]]], [qa17, 0, [[0, X17]]]], [step, step], qa17, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([017], 2, [[q017, 0, [[0, 017]]], [q117, 1, [[0, X17]]], [qa17, 0, [[0, X17]]]], [step, step], qa17, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0043/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 44. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q018, q118, qa18, qr18], input_alphabet=[018], tape_alphabet=[018, X18, _18], blank=_18, start=q018, accept=qa18, reject=qr18, transitions=[[q018, 018, q118, X18, R], [q018, _18, qr18, _18, S], [q118, _18, qa18, _18, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[018], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([018], 2, [[q018, 0, [[0, 018]]], [q118, 1, [[0, X18]]], [qa18, 0, [[0, X18]]]], [step, step], qa18, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([018], 2, [[q018, 0, [[0, 018]]], [q118, 1, [[0, X18]]], [qa18, 0, [[0, X18]]]], [step, step], qa18, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0044/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 45. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q019, q119, qa19, qr19], input_alphabet=[019], tape_alphabet=[019, X19, _19], blank=_19, start=q019, accept=qa19, reject=qr19, transitions=[[q019, 019, q119, X19, R], [q019, _19, qr19, _19, S], [q119, _19, qa19, _19, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[019], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([019], 2, [[q019, 0, [[0, 019]]], [q119, 1, [[0, X19]]], [qa19, 0, [[0, X19]]]], [step, step], qa19, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([019], 2, [[q019, 0, [[0, 019]]], [q119, 1, [[0, X19]]], [qa19, 0, [[0, X19]]]], [step, step], qa19, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0045/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 46. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q020, q120, qa20, qr20], input_alphabet=[020], tape_alphabet=[020, X20, _20], blank=_20, start=q020, accept=qa20, reject=qr20, transitions=[[q020, 020, q120, X20, R], [q020, _20, qr20, _20, S], [q120, _20, qa20, _20, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[020], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([020], 2, [[q020, 0, [[0, 020]]], [q120, 1, [[0, X20]]], [qa20, 0, [[0, X20]]]], [step, step], qa20, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([020], 2, [[q020, 0, [[0, 020]]], [q120, 1, [[0, X20]]], [qa20, 0, [[0, X20]]]], [step, step], qa20, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0046/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 47. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q021, q121, qa21, qr21], input_alphabet=[021], tape_alphabet=[021, X21, _21], blank=_21, start=q021, accept=qa21, reject=qr21, transitions=[[q021, 021, q121, X21, R], [q021, _21, qr21, _21, S], [q121, _21, qa21, _21, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[021], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([021], 2, [[q021, 0, [[0, 021]]], [q121, 1, [[0, X21]]], [qa21, 0, [[0, X21]]]], [step, step], qa21, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([021], 2, [[q021, 0, [[0, 021]]], [q121, 1, [[0, X21]]], [qa21, 0, [[0, X21]]]], [step, step], qa21, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0047/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 48. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q022, q122, qa22, qr22], input_alphabet=[022], tape_alphabet=[022, X22, _22], blank=_22, start=q022, accept=qa22, reject=qr22, transitions=[[q022, 022, q122, X22, R], [q022, _22, qr22, _22, S], [q122, _22, qa22, _22, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[022], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([022], 2, [[q022, 0, [[0, 022]]], [q122, 1, [[0, X22]]], [qa22, 0, [[0, X22]]]], [step, step], qa22, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([022], 2, [[q022, 0, [[0, 022]]], [q122, 1, [[0, X22]]], [qa22, 0, [[0, X22]]]], [step, step], qa22, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0048/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 49. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q023, q123, qa23, qr23], input_alphabet=[023], tape_alphabet=[023, X23, _23], blank=_23, start=q023, accept=qa23, reject=qr23, transitions=[[q023, 023, q123, X23, R], [q023, _23, qr23, _23, S], [q123, _23, qa23, _23, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[023], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([023], 2, [[q023, 0, [[0, 023]]], [q123, 1, [[0, X23]]], [qa23, 0, [[0, X23]]]], [step, step], qa23, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([023], 2, [[q023, 0, [[0, 023]]], [q123, 1, [[0, X23]]], [qa23, 0, [[0, X23]]]], [step, step], qa23, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0049/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 50. A Finite Configuration Trace Has Only Its Declared Input and Budget Scope

Use deterministic single-tape states=[q024, q124, qa24, qr24], input_alphabet=[024], tape_alphabet=[024, X24, _24], blank=_24, start=q024, accept=qa24, reject=qr24, transitions=[[q024, 024, q124, X24, R], [q024, _24, qr24, _24, S], [q124, _24, qa24, _24, L]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Input=[024], step budget=2. Give (input, step_budget, configurations, step_outcomes, final_state, local_outcome, global_decider_certificate, boundary).

stuck means no transition for the current read symbol, not automatic rejection. Failure to halt within a budget is not proof of an infinite loop.

View answer and explanation([024], 2, [[q024, 0, [[0, 024]]], [q124, 1, [[0, X24]]], [qa24, 0, [[0, X24]]]], [step, step], qa24, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider)

Start at head0 with input in cells0 onward. Apply write/state/move exactly, deleting blank entries from the sparse record; stop on accept/reject, budget exhaustion or a missing rule. Missing rules produce stuck with the unchanged configuration, not automatic reject. The fixed No withholds a global decider certificate; a finite nonhalt is not an infinite-loop proof. The exact result, retaining the requested order, is ([024], 2, [[q024, 0, [[0, 024]]], [q124, 1, [[0, X24]]], [qa24, 0, [[0, X24]]]], [step, step], qa24, accept, No, finite_trace_is_input_local_and_does_not_certify_a_global_decider).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0050/)

Turing-machine decision model · qf-mod44-tm-exact-step-trace · candidate





## 51. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a0, b0], tape_alphabet=[a0, b0, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a0, scan, a0, R], [scan, b0, accept, b0, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b0. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a0, b0], [_, a0, b0], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a0, scan, a0, R], [scan, b0, accept, b0, S]], b0, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a0, b0], [_, a0, b0], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a0, scan, a0, R], [scan, b0, accept, b0, S]], b0, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0051/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 52. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a1, b1], tape_alphabet=[a1, b1, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a1, scan, a1, R], [scan, b1, accept, b1, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b1. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a1, b1], [_, a1, b1], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a1, scan, a1, R], [scan, b1, accept, b1, S]], b1, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a1, b1], [_, a1, b1], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a1, scan, a1, R], [scan, b1, accept, b1, S]], b1, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0052/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 53. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a2, b2], tape_alphabet=[a2, b2, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a2, scan, a2, R], [scan, b2, accept, b2, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b2. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a2, b2], [_, a2, b2], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a2, scan, a2, R], [scan, b2, accept, b2, S]], b2, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a2, b2], [_, a2, b2], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a2, scan, a2, R], [scan, b2, accept, b2, S]], b2, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0053/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 54. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a3, b3], tape_alphabet=[a3, b3, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a3, scan, a3, R], [scan, b3, accept, b3, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b3. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a3, b3], [_, a3, b3], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a3, scan, a3, R], [scan, b3, accept, b3, S]], b3, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a3, b3], [_, a3, b3], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a3, scan, a3, R], [scan, b3, accept, b3, S]], b3, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0054/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 55. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a4, b4], tape_alphabet=[a4, b4, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a4, scan, a4, R], [scan, b4, accept, b4, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b4. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a4, b4], [_, a4, b4], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a4, scan, a4, R], [scan, b4, accept, b4, S]], b4, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a4, b4], [_, a4, b4], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a4, scan, a4, R], [scan, b4, accept, b4, S]], b4, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0055/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 56. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a5, b5], tape_alphabet=[a5, b5, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a5, scan, a5, R], [scan, b5, accept, b5, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b5. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a5, b5], [_, a5, b5], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a5, scan, a5, R], [scan, b5, accept, b5, S]], b5, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a5, b5], [_, a5, b5], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a5, scan, a5, R], [scan, b5, accept, b5, S]], b5, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0056/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 57. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a6, b6], tape_alphabet=[a6, b6, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a6, scan, a6, R], [scan, b6, accept, b6, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b6. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a6, b6], [_, a6, b6], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a6, scan, a6, R], [scan, b6, accept, b6, S]], b6, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a6, b6], [_, a6, b6], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a6, scan, a6, R], [scan, b6, accept, b6, S]], b6, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0057/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 58. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a7, b7], tape_alphabet=[a7, b7, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a7, scan, a7, R], [scan, b7, accept, b7, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b7. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a7, b7], [_, a7, b7], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a7, scan, a7, R], [scan, b7, accept, b7, S]], b7, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a7, b7], [_, a7, b7], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a7, scan, a7, R], [scan, b7, accept, b7, S]], b7, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0058/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 59. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a8, b8], tape_alphabet=[a8, b8, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a8, scan, a8, R], [scan, b8, accept, b8, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b8. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a8, b8], [_, a8, b8], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a8, scan, a8, R], [scan, b8, accept, b8, S]], b8, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a8, b8], [_, a8, b8], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a8, scan, a8, R], [scan, b8, accept, b8, S]], b8, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0059/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 60. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a9, b9], tape_alphabet=[a9, b9, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a9, scan, a9, R], [scan, b9, accept, b9, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b9. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a9, b9], [_, a9, b9], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a9, scan, a9, R], [scan, b9, accept, b9, S]], b9, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a9, b9], [_, a9, b9], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a9, scan, a9, R], [scan, b9, accept, b9, S]], b9, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0060/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 61. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a10, b10], tape_alphabet=[a10, b10, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a10, scan, a10, R], [scan, b10, accept, b10, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b10. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a10, b10], [_, a10, b10], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a10, scan, a10, R], [scan, b10, accept, b10, S]], b10, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a10, b10], [_, a10, b10], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a10, scan, a10, R], [scan, b10, accept, b10, S]], b10, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0061/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 62. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a11, b11], tape_alphabet=[a11, b11, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a11, scan, a11, R], [scan, b11, accept, b11, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b11. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a11, b11], [_, a11, b11], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a11, scan, a11, R], [scan, b11, accept, b11, S]], b11, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a11, b11], [_, a11, b11], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a11, scan, a11, R], [scan, b11, accept, b11, S]], b11, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0062/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 63. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a12, b12], tape_alphabet=[a12, b12, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a12, scan, a12, R], [scan, b12, accept, b12, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b12. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a12, b12], [_, a12, b12], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a12, scan, a12, R], [scan, b12, accept, b12, S]], b12, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a12, b12], [_, a12, b12], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a12, scan, a12, R], [scan, b12, accept, b12, S]], b12, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0063/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 64. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a13, b13], tape_alphabet=[a13, b13, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a13, scan, a13, R], [scan, b13, accept, b13, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b13. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a13, b13], [_, a13, b13], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a13, scan, a13, R], [scan, b13, accept, b13, S]], b13, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a13, b13], [_, a13, b13], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a13, scan, a13, R], [scan, b13, accept, b13, S]], b13, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0064/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 65. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a14, b14], tape_alphabet=[a14, b14, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a14, scan, a14, R], [scan, b14, accept, b14, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b14. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a14, b14], [_, a14, b14], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a14, scan, a14, R], [scan, b14, accept, b14, S]], b14, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a14, b14], [_, a14, b14], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a14, scan, a14, R], [scan, b14, accept, b14, S]], b14, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0065/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 66. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a15, b15], tape_alphabet=[a15, b15, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a15, scan, a15, R], [scan, b15, accept, b15, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b15. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a15, b15], [_, a15, b15], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a15, scan, a15, R], [scan, b15, accept, b15, S]], b15, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a15, b15], [_, a15, b15], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a15, scan, a15, R], [scan, b15, accept, b15, S]], b15, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0066/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 67. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a16, b16], tape_alphabet=[a16, b16, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a16, scan, a16, R], [scan, b16, accept, b16, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b16. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a16, b16], [_, a16, b16], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a16, scan, a16, R], [scan, b16, accept, b16, S]], b16, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a16, b16], [_, a16, b16], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a16, scan, a16, R], [scan, b16, accept, b16, S]], b16, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0067/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 68. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a17, b17], tape_alphabet=[a17, b17, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a17, scan, a17, R], [scan, b17, accept, b17, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b17. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a17, b17], [_, a17, b17], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a17, scan, a17, R], [scan, b17, accept, b17, S]], b17, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a17, b17], [_, a17, b17], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a17, scan, a17, R], [scan, b17, accept, b17, S]], b17, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0068/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 69. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a18, b18], tape_alphabet=[a18, b18, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a18, scan, a18, R], [scan, b18, accept, b18, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b18. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a18, b18], [_, a18, b18], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a18, scan, a18, R], [scan, b18, accept, b18, S]], b18, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a18, b18], [_, a18, b18], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a18, scan, a18, R], [scan, b18, accept, b18, S]], b18, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0069/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 70. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a19, b19], tape_alphabet=[a19, b19, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a19, scan, a19, R], [scan, b19, accept, b19, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b19. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a19, b19], [_, a19, b19], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a19, scan, a19, R], [scan, b19, accept, b19, S]], b19, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a19, b19], [_, a19, b19], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a19, scan, a19, R], [scan, b19, accept, b19, S]], b19, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0070/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 71. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a20, b20], tape_alphabet=[a20, b20, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a20, scan, a20, R], [scan, b20, accept, b20, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b20. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a20, b20], [_, a20, b20], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a20, scan, a20, R], [scan, b20, accept, b20, S]], b20, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a20, b20], [_, a20, b20], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a20, scan, a20, R], [scan, b20, accept, b20, S]], b20, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0071/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 72. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a21, b21], tape_alphabet=[a21, b21, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a21, scan, a21, R], [scan, b21, accept, b21, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b21. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a21, b21], [_, a21, b21], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a21, scan, a21, R], [scan, b21, accept, b21, S]], b21, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a21, b21], [_, a21, b21], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a21, scan, a21, R], [scan, b21, accept, b21, S]], b21, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0072/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 73. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a22, b22], tape_alphabet=[a22, b22, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a22, scan, a22, R], [scan, b22, accept, b22, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b22. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a22, b22], [_, a22, b22], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a22, scan, a22, R], [scan, b22, accept, b22, S]], b22, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a22, b22], [_, a22, b22], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a22, scan, a22, R], [scan, b22, accept, b22, S]], b22, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0073/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 74. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a23, b23], tape_alphabet=[a23, b23, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a23, scan, a23, R], [scan, b23, accept, b23, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b23. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a23, b23], [_, a23, b23], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a23, scan, a23, R], [scan, b23, accept, b23, S]], b23, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a23, b23], [_, a23, b23], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a23, scan, a23, R], [scan, b23, accept, b23, S]], b23, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0074/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 75. A Canonical Scan Schema Has an All-Input Ranking and Correct Answers

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a24, b24], tape_alphabet=[a24, b24, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a24, scan, a24, R], [scan, b24, accept, b24, S], [scan, _, reject, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target input symbol=b24. Give (states, input_alphabet, tape_alphabet, blank, start, accept, reject, transitions, target_symbol, ranking, decrease_rule, step_bound, schema_decider_certificate, certificate).

View answer and explanation([accept, reject, scan], [a24, b24], [_, a24, b24], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a24, scan, a24, R], [scan, b24, accept, b24, S]], b24, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words)

Require the exact canonical schema: copy and move right on nontarget symbols, accept on the target, reject on blank. For every legal word w, unscanned input cells plus1 decreases on scanning; assign rank0 after halting, including the final accept/reject step. The bound is |w|+1 and the answer is correct for the language containing the target. This pattern-bound certificate does not certify arbitrary machines from sample traces. The exact result, retaining the requested order, is ([accept, reject, scan], [a24, b24], [_, a24, b24], _, scan, accept, reject, [[scan, _, reject, _, S], [scan, a24, scan, a24, R], [scan, b24, accept, b24, S]], b24, ranking=unscanned_input_cells_plus_one, strictly_decreases_on_every_scan_step, |w|+1, Yes, global_termination_and_answer_correctness_certificate_for_all_words).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0075/)

Turing-machine decision model · qf-mod44-tm-global-decider-ranking · candidate





## 76. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a0, b0], tape_alphabet=[a0, b0, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a0, scan, a0, R], [scan, b0, accept, b0, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b0, legal target-free input=[a0, a0]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a0, b0], [_, a0, b0], [[scan, _, scan, _, S], [scan, a0, scan, a0, R], [scan, b0, accept, b0, S]], b0, [a0, a0], [scan, 2, [[0, a0], [1, a0]]], [scan, 2, [[0, a0], [1, a0]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a0, b0], [_, a0, b0], [[scan, _, scan, _, S], [scan, a0, scan, a0, R], [scan, b0, accept, b0, S]], b0, [a0, a0], [scan, 2, [[0, a0], [1, a0]]], [scan, 2, [[0, a0], [1, a0]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0076/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 77. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a1, b1], tape_alphabet=[a1, b1, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a1, scan, a1, R], [scan, b1, accept, b1, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b1, legal target-free input=[a1, a1]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a1, b1], [_, a1, b1], [[scan, _, scan, _, S], [scan, a1, scan, a1, R], [scan, b1, accept, b1, S]], b1, [a1, a1], [scan, 2, [[0, a1], [1, a1]]], [scan, 2, [[0, a1], [1, a1]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a1, b1], [_, a1, b1], [[scan, _, scan, _, S], [scan, a1, scan, a1, R], [scan, b1, accept, b1, S]], b1, [a1, a1], [scan, 2, [[0, a1], [1, a1]]], [scan, 2, [[0, a1], [1, a1]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0077/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 78. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a2, b2], tape_alphabet=[a2, b2, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a2, scan, a2, R], [scan, b2, accept, b2, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b2, legal target-free input=[a2, a2]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a2, b2], [_, a2, b2], [[scan, _, scan, _, S], [scan, a2, scan, a2, R], [scan, b2, accept, b2, S]], b2, [a2, a2], [scan, 2, [[0, a2], [1, a2]]], [scan, 2, [[0, a2], [1, a2]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a2, b2], [_, a2, b2], [[scan, _, scan, _, S], [scan, a2, scan, a2, R], [scan, b2, accept, b2, S]], b2, [a2, a2], [scan, 2, [[0, a2], [1, a2]]], [scan, 2, [[0, a2], [1, a2]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0078/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 79. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a3, b3], tape_alphabet=[a3, b3, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a3, scan, a3, R], [scan, b3, accept, b3, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b3, legal target-free input=[a3, a3]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a3, b3], [_, a3, b3], [[scan, _, scan, _, S], [scan, a3, scan, a3, R], [scan, b3, accept, b3, S]], b3, [a3, a3], [scan, 2, [[0, a3], [1, a3]]], [scan, 2, [[0, a3], [1, a3]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a3, b3], [_, a3, b3], [[scan, _, scan, _, S], [scan, a3, scan, a3, R], [scan, b3, accept, b3, S]], b3, [a3, a3], [scan, 2, [[0, a3], [1, a3]]], [scan, 2, [[0, a3], [1, a3]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0079/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 80. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a4, b4], tape_alphabet=[a4, b4, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a4, scan, a4, R], [scan, b4, accept, b4, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b4, legal target-free input=[a4, a4]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a4, b4], [_, a4, b4], [[scan, _, scan, _, S], [scan, a4, scan, a4, R], [scan, b4, accept, b4, S]], b4, [a4, a4], [scan, 2, [[0, a4], [1, a4]]], [scan, 2, [[0, a4], [1, a4]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a4, b4], [_, a4, b4], [[scan, _, scan, _, S], [scan, a4, scan, a4, R], [scan, b4, accept, b4, S]], b4, [a4, a4], [scan, 2, [[0, a4], [1, a4]]], [scan, 2, [[0, a4], [1, a4]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0080/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 81. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a5, b5], tape_alphabet=[a5, b5, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a5, scan, a5, R], [scan, b5, accept, b5, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b5, legal target-free input=[a5, a5]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a5, b5], [_, a5, b5], [[scan, _, scan, _, S], [scan, a5, scan, a5, R], [scan, b5, accept, b5, S]], b5, [a5, a5], [scan, 2, [[0, a5], [1, a5]]], [scan, 2, [[0, a5], [1, a5]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a5, b5], [_, a5, b5], [[scan, _, scan, _, S], [scan, a5, scan, a5, R], [scan, b5, accept, b5, S]], b5, [a5, a5], [scan, 2, [[0, a5], [1, a5]]], [scan, 2, [[0, a5], [1, a5]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0081/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 82. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a6, b6], tape_alphabet=[a6, b6, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a6, scan, a6, R], [scan, b6, accept, b6, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b6, legal target-free input=[a6, a6]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a6, b6], [_, a6, b6], [[scan, _, scan, _, S], [scan, a6, scan, a6, R], [scan, b6, accept, b6, S]], b6, [a6, a6], [scan, 2, [[0, a6], [1, a6]]], [scan, 2, [[0, a6], [1, a6]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a6, b6], [_, a6, b6], [[scan, _, scan, _, S], [scan, a6, scan, a6, R], [scan, b6, accept, b6, S]], b6, [a6, a6], [scan, 2, [[0, a6], [1, a6]]], [scan, 2, [[0, a6], [1, a6]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0082/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 83. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a7, b7], tape_alphabet=[a7, b7, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a7, scan, a7, R], [scan, b7, accept, b7, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b7, legal target-free input=[a7, a7]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a7, b7], [_, a7, b7], [[scan, _, scan, _, S], [scan, a7, scan, a7, R], [scan, b7, accept, b7, S]], b7, [a7, a7], [scan, 2, [[0, a7], [1, a7]]], [scan, 2, [[0, a7], [1, a7]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a7, b7], [_, a7, b7], [[scan, _, scan, _, S], [scan, a7, scan, a7, R], [scan, b7, accept, b7, S]], b7, [a7, a7], [scan, 2, [[0, a7], [1, a7]]], [scan, 2, [[0, a7], [1, a7]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0083/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 84. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a8, b8], tape_alphabet=[a8, b8, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a8, scan, a8, R], [scan, b8, accept, b8, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b8, legal target-free input=[a8, a8]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a8, b8], [_, a8, b8], [[scan, _, scan, _, S], [scan, a8, scan, a8, R], [scan, b8, accept, b8, S]], b8, [a8, a8], [scan, 2, [[0, a8], [1, a8]]], [scan, 2, [[0, a8], [1, a8]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a8, b8], [_, a8, b8], [[scan, _, scan, _, S], [scan, a8, scan, a8, R], [scan, b8, accept, b8, S]], b8, [a8, a8], [scan, 2, [[0, a8], [1, a8]]], [scan, 2, [[0, a8], [1, a8]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0084/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 85. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a9, b9], tape_alphabet=[a9, b9, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a9, scan, a9, R], [scan, b9, accept, b9, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b9, legal target-free input=[a9, a9]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a9, b9], [_, a9, b9], [[scan, _, scan, _, S], [scan, a9, scan, a9, R], [scan, b9, accept, b9, S]], b9, [a9, a9], [scan, 2, [[0, a9], [1, a9]]], [scan, 2, [[0, a9], [1, a9]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a9, b9], [_, a9, b9], [[scan, _, scan, _, S], [scan, a9, scan, a9, R], [scan, b9, accept, b9, S]], b9, [a9, a9], [scan, 2, [[0, a9], [1, a9]]], [scan, 2, [[0, a9], [1, a9]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0085/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 86. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a10, b10], tape_alphabet=[a10, b10, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a10, scan, a10, R], [scan, b10, accept, b10, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b10, legal target-free input=[a10, a10]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a10, b10], [_, a10, b10], [[scan, _, scan, _, S], [scan, a10, scan, a10, R], [scan, b10, accept, b10, S]], b10, [a10, a10], [scan, 2, [[0, a10], [1, a10]]], [scan, 2, [[0, a10], [1, a10]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a10, b10], [_, a10, b10], [[scan, _, scan, _, S], [scan, a10, scan, a10, R], [scan, b10, accept, b10, S]], b10, [a10, a10], [scan, 2, [[0, a10], [1, a10]]], [scan, 2, [[0, a10], [1, a10]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0086/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 87. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a11, b11], tape_alphabet=[a11, b11, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a11, scan, a11, R], [scan, b11, accept, b11, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b11, legal target-free input=[a11, a11]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a11, b11], [_, a11, b11], [[scan, _, scan, _, S], [scan, a11, scan, a11, R], [scan, b11, accept, b11, S]], b11, [a11, a11], [scan, 2, [[0, a11], [1, a11]]], [scan, 2, [[0, a11], [1, a11]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a11, b11], [_, a11, b11], [[scan, _, scan, _, S], [scan, a11, scan, a11, R], [scan, b11, accept, b11, S]], b11, [a11, a11], [scan, 2, [[0, a11], [1, a11]]], [scan, 2, [[0, a11], [1, a11]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0087/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 88. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a12, b12], tape_alphabet=[a12, b12, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a12, scan, a12, R], [scan, b12, accept, b12, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b12, legal target-free input=[a12, a12]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a12, b12], [_, a12, b12], [[scan, _, scan, _, S], [scan, a12, scan, a12, R], [scan, b12, accept, b12, S]], b12, [a12, a12], [scan, 2, [[0, a12], [1, a12]]], [scan, 2, [[0, a12], [1, a12]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a12, b12], [_, a12, b12], [[scan, _, scan, _, S], [scan, a12, scan, a12, R], [scan, b12, accept, b12, S]], b12, [a12, a12], [scan, 2, [[0, a12], [1, a12]]], [scan, 2, [[0, a12], [1, a12]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0088/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 89. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a13, b13], tape_alphabet=[a13, b13, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a13, scan, a13, R], [scan, b13, accept, b13, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b13, legal target-free input=[a13, a13]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a13, b13], [_, a13, b13], [[scan, _, scan, _, S], [scan, a13, scan, a13, R], [scan, b13, accept, b13, S]], b13, [a13, a13], [scan, 2, [[0, a13], [1, a13]]], [scan, 2, [[0, a13], [1, a13]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a13, b13], [_, a13, b13], [[scan, _, scan, _, S], [scan, a13, scan, a13, R], [scan, b13, accept, b13, S]], b13, [a13, a13], [scan, 2, [[0, a13], [1, a13]]], [scan, 2, [[0, a13], [1, a13]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0089/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 90. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a14, b14], tape_alphabet=[a14, b14, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a14, scan, a14, R], [scan, b14, accept, b14, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b14, legal target-free input=[a14, a14]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a14, b14], [_, a14, b14], [[scan, _, scan, _, S], [scan, a14, scan, a14, R], [scan, b14, accept, b14, S]], b14, [a14, a14], [scan, 2, [[0, a14], [1, a14]]], [scan, 2, [[0, a14], [1, a14]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a14, b14], [_, a14, b14], [[scan, _, scan, _, S], [scan, a14, scan, a14, R], [scan, b14, accept, b14, S]], b14, [a14, a14], [scan, 2, [[0, a14], [1, a14]]], [scan, 2, [[0, a14], [1, a14]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0090/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 91. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a15, b15], tape_alphabet=[a15, b15, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a15, scan, a15, R], [scan, b15, accept, b15, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b15, legal target-free input=[a15, a15]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a15, b15], [_, a15, b15], [[scan, _, scan, _, S], [scan, a15, scan, a15, R], [scan, b15, accept, b15, S]], b15, [a15, a15], [scan, 2, [[0, a15], [1, a15]]], [scan, 2, [[0, a15], [1, a15]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a15, b15], [_, a15, b15], [[scan, _, scan, _, S], [scan, a15, scan, a15, R], [scan, b15, accept, b15, S]], b15, [a15, a15], [scan, 2, [[0, a15], [1, a15]]], [scan, 2, [[0, a15], [1, a15]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0091/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 92. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a16, b16], tape_alphabet=[a16, b16, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a16, scan, a16, R], [scan, b16, accept, b16, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b16, legal target-free input=[a16, a16]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a16, b16], [_, a16, b16], [[scan, _, scan, _, S], [scan, a16, scan, a16, R], [scan, b16, accept, b16, S]], b16, [a16, a16], [scan, 2, [[0, a16], [1, a16]]], [scan, 2, [[0, a16], [1, a16]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a16, b16], [_, a16, b16], [[scan, _, scan, _, S], [scan, a16, scan, a16, R], [scan, b16, accept, b16, S]], b16, [a16, a16], [scan, 2, [[0, a16], [1, a16]]], [scan, 2, [[0, a16], [1, a16]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0092/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 93. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a17, b17], tape_alphabet=[a17, b17, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a17, scan, a17, R], [scan, b17, accept, b17, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b17, legal target-free input=[a17, a17]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a17, b17], [_, a17, b17], [[scan, _, scan, _, S], [scan, a17, scan, a17, R], [scan, b17, accept, b17, S]], b17, [a17, a17], [scan, 2, [[0, a17], [1, a17]]], [scan, 2, [[0, a17], [1, a17]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a17, b17], [_, a17, b17], [[scan, _, scan, _, S], [scan, a17, scan, a17, R], [scan, b17, accept, b17, S]], b17, [a17, a17], [scan, 2, [[0, a17], [1, a17]]], [scan, 2, [[0, a17], [1, a17]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0093/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 94. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a18, b18], tape_alphabet=[a18, b18, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a18, scan, a18, R], [scan, b18, accept, b18, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b18, legal target-free input=[a18, a18]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a18, b18], [_, a18, b18], [[scan, _, scan, _, S], [scan, a18, scan, a18, R], [scan, b18, accept, b18, S]], b18, [a18, a18], [scan, 2, [[0, a18], [1, a18]]], [scan, 2, [[0, a18], [1, a18]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a18, b18], [_, a18, b18], [[scan, _, scan, _, S], [scan, a18, scan, a18, R], [scan, b18, accept, b18, S]], b18, [a18, a18], [scan, 2, [[0, a18], [1, a18]]], [scan, 2, [[0, a18], [1, a18]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0094/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 95. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a19, b19], tape_alphabet=[a19, b19, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a19, scan, a19, R], [scan, b19, accept, b19, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b19, legal target-free input=[a19, a19]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a19, b19], [_, a19, b19], [[scan, _, scan, _, S], [scan, a19, scan, a19, R], [scan, b19, accept, b19, S]], b19, [a19, a19], [scan, 2, [[0, a19], [1, a19]]], [scan, 2, [[0, a19], [1, a19]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a19, b19], [_, a19, b19], [[scan, _, scan, _, S], [scan, a19, scan, a19, R], [scan, b19, accept, b19, S]], b19, [a19, a19], [scan, 2, [[0, a19], [1, a19]]], [scan, 2, [[0, a19], [1, a19]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0095/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 96. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a20, b20], tape_alphabet=[a20, b20, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a20, scan, a20, R], [scan, b20, accept, b20, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b20, legal target-free input=[a20, a20]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a20, b20], [_, a20, b20], [[scan, _, scan, _, S], [scan, a20, scan, a20, R], [scan, b20, accept, b20, S]], b20, [a20, a20], [scan, 2, [[0, a20], [1, a20]]], [scan, 2, [[0, a20], [1, a20]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a20, b20], [_, a20, b20], [[scan, _, scan, _, S], [scan, a20, scan, a20, R], [scan, b20, accept, b20, S]], b20, [a20, a20], [scan, 2, [[0, a20], [1, a20]]], [scan, 2, [[0, a20], [1, a20]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0096/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 97. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a21, b21], tape_alphabet=[a21, b21, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a21, scan, a21, R], [scan, b21, accept, b21, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b21, legal target-free input=[a21, a21]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a21, b21], [_, a21, b21], [[scan, _, scan, _, S], [scan, a21, scan, a21, R], [scan, b21, accept, b21, S]], b21, [a21, a21], [scan, 2, [[0, a21], [1, a21]]], [scan, 2, [[0, a21], [1, a21]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a21, b21], [_, a21, b21], [[scan, _, scan, _, S], [scan, a21, scan, a21, R], [scan, b21, accept, b21, S]], b21, [a21, a21], [scan, 2, [[0, a21], [1, a21]]], [scan, 2, [[0, a21], [1, a21]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0097/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 98. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a22, b22], tape_alphabet=[a22, b22, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a22, scan, a22, R], [scan, b22, accept, b22, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b22, legal target-free input=[a22, a22]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a22, b22], [_, a22, b22], [[scan, _, scan, _, S], [scan, a22, scan, a22, R], [scan, b22, accept, b22, S]], b22, [a22, a22], [scan, 2, [[0, a22], [1, a22]]], [scan, 2, [[0, a22], [1, a22]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a22, b22], [_, a22, b22], [[scan, _, scan, _, S], [scan, a22, scan, a22, R], [scan, b22, accept, b22, S]], b22, [a22, a22], [scan, 2, [[0, a22], [1, a22]]], [scan, 2, [[0, a22], [1, a22]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0098/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 99. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a23, b23], tape_alphabet=[a23, b23, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a23, scan, a23, R], [scan, b23, accept, b23, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b23, legal target-free input=[a23, a23]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a23, b23], [_, a23, b23], [[scan, _, scan, _, S], [scan, a23, scan, a23, R], [scan, b23, accept, b23, S]], b23, [a23, a23], [scan, 2, [[0, a23], [1, a23]]], [scan, 2, [[0, a23], [1, a23]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a23, b23], [_, a23, b23], [[scan, _, scan, _, S], [scan, a23, scan, a23, R], [scan, b23, accept, b23, S]], b23, [a23, a23], [scan, 2, [[0, a23], [1, a23]]], [scan, 2, [[0, a23], [1, a23]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0099/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate





## 100. An Explicit Repeated Blank Configuration Proves This Recognizer Is Not a Decider

Use deterministic single-tape states=[scan, accept, reject], input_alphabet=[a24, b24], tape_alphabet=[a24, b24, _], blank=_, start=scan, accept=accept, reject=reject, transitions=[[scan, a24, scan, a24, R], [scan, b24, accept, b24, S], [scan, _, scan, _, S]]. Each transition is [state,read,next,write,L/R/S]; keys [state,read] are unique and halting states have no outgoing transitions. Input symbols form a blank-free tape subset. The integer-indexed tape is blank at every unlisted cell, with configuration [state,head,sparse nonblank tape]; L may move left of0. Canonical answer transitions/tape cells retain the source sort convention. Target=b24, legal target-free input=[a24, a24]. Give (states, input_alphabet, tape_alphabet, transitions, target_symbol, input, loop_configuration, next_configuration, exact_self_loop, decider, boundary).

View answer and explanation([accept, reject, scan], [a24, b24], [_, a24, b24], [[scan, _, scan, _, S], [scan, a24, scan, a24, R], [scan, b24, accept, b24, S]], b24, [a24, a24], [scan, 2, [[0, a24], [1, a24]]], [scan, 2, [[0, a24], [1, a24]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input)

Require the canonical recognizer schema that loops in scan on blank without moving or changing tape. After all nontarget input cells are scanned, the exact blank configuration repeats forever. This is a constructive nontermination witness for that input, stronger than bounded simulation failure; it does not claim every recognizer fails to be a decider. The exact result, retaining the requested order, is ([accept, reject, scan], [a24, b24], [_, a24, b24], [[scan, _, scan, _, S], [scan, a24, scan, a24, R], [scan, b24, accept, b24, S]], b24, [a24, a24], [scan, 2, [[0, a24], [1, a24]]], [scan, 2, [[0, a24], [1, a24]]], Yes, No, explicit_blank_self_loop_proves_this_recognizer_is_not_a_decider_on_the_target_free_input).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261113-r05-turing-machine-decision-model-0100/)

Turing-machine decision model · qf-mod44-tm-recognizer-loop-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261113-r05-turing-machine-decision-model-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261113-r05-turing-machine-decision-model-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261113-r05-turing-machine-decision-model-foundations/validation-report.json)