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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_11755::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_11755", "language": "French", "seed_prompt": "I came across Norton's dome and I don't agree that it proves anything.\nFirst, here's an obviously ridiculous and completely nonsensical example that I constructed from thinking about simple harmonic motion:\nConsider a particle of unit mass at rest in free space, at $x = 0$. Newton's second law states that $\\ddot{x} = 0$ at $t=0$. There's an obvious way to describe the particle's motion that satisfies Newton's second law of motion, and that is $x(t) = 0$. Let us denote this solution by $s_1(t)$.\nHowever, here's another non-obvious way to describe the particle's motion that satisfies Newton's second law:\n$$\nx(t) = \\begin{cases}\n\\sin(t - T) - t + T \\quad &\\text{if }\\ t \\ge T \\\\\n0 & \\text{if }\\ t < T\n\\end{cases}\n$$\nLet us denote this solution by $s_2(t)$.\nNotice that $s_2(t)$, $\\dot{s_2}(t)$ and $\\ddot{s_2}(t)$ all obviously equal $0$ for all $t < T$.\nNow consider time $t = T$. Notice that:\n$$\n\\begin{align*}\ns_2(T) & = \\sin(T - T) - T + T = 0 \\\\\n\\dot{s_2}(T) & = \\cos(T - T) - 1 = 0 \\\\\n\\ddot{s_2}(T) & = -\\sin(T - T) = 0\n\\end{align*}\n$$\nThus, $s_2(t)$ is a completely valid solution which obeys the constraint that the particle is at rest at time $t=0$. However, notice that as $t$ grows beyond $T$, the particle starts to spontaneously shoot off to infinity in the negative $x$ direction. Also notice that $T$ is an arbitrarily chosen constant, it could assume any positive real number as its value.\nThus, the second solution states that the particle could spontaneously start shooting off to infinity in the negative $x$ direction after some arbitrary amount of time $T$.\nSo we have proven that Newtonian mechanics isn't deterministic, can harbor uncaused events, yada yada.\nWhy are Norton's dome and my example not valid?\nThey're incorrect because in those solutions, the particle isn't really at rest when the spontaneous motion starts occurring. Notice that the fourth derivative of $r(t) = \\frac{d^4}{dt^4} \\left( \\frac{(t - T)^4}{144} \\right) = \\frac{1}{6}$, and that the third derivative of $s_2(t)$ at $t=T$ is equal to $-\\cos(T-T) = -1$.\nClearly Norton believes that an object being \"at rest\" does not necessitate that $\\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n \\ge 2 $. He believes that Newton's first law does not imply $ \\frac{d^2 \\vec{p}}{dt^2} \\implies \\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n \\ge 2 $.\nHowever, believing this leads to all sorts of completely nonsensical predictions for even the simplest system one can think of, a particle at rest in free space, as I have shown above.\nTherefore, it only makes sense to believe that $ \\frac{d^2 \\vec{p}}{dt^2} = \\vec{0} \\implies \\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n > 2$.\nSo, coming to my questions: Are my arguments valid? Is Norton's second solution just plain incorrect? If you disagree with my arguments and instead agree with Norton, what do you think of my completely nonsensical second solution for a particle at rest in free space?\n", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "strong_b", "aspects": {"naturalness": "strong_b", "cultural_appropriateness": "tie_both_bad", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "smith", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:23:46", "duration_ms": 290119}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_40351::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_40351", "language": "French", "seed_prompt": "Benzenediazonium tetrafluoroborate decomposes in dimethyl sulfoxide with evolution of nitrogen and phenylation of any aromatic present. The following isomer ratios were obtained for the phenylation of toluene, chlorobenzene, and nitrobenzene.\\n\\n| Compound | \\( \\text{p/o} \\) | \\( \\text{m/l} \\) | \\( \\text{o/m} \\) |\\n| -------- | ---------------- | ---------------- | ---------------- |\\n| PhCH\\(_3\\) | 47:3 | 2:1·1 | 31·6 |\\n| PhCl | 48 | 2·40 | 27·8 |\\n| PhNO\\(_2\\) | 38·1 | 48·4 | 13·5 |\\n\\n## Miscellaneous Problems\\n\\nThe rates of reaction correlate with \\(\\sigma^*\\) and the value of \\(\\rho\\) is −0.74. The isomer ratios are the same for anisole-d\\(_6\\) and anisole. Is phenylunion under these conditions free radical or electrophilic and what is the rate-determining step?", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "strong_a", "aspects": {"naturalness": "strong_a", "cultural_appropriateness": "tie_both_bad", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "teacher", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:26:32", "duration_ms": 166051}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_42290::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_42290", "language": "French", "seed_prompt": "Suppose we have a spin $\\frac{1}{2}$ particle in the spin-up state along the $z$-axis, $\\lvert \\uparrow \\rangle$, and after $t$ seconds of evolution under the Schrodinger equation it is in state $\\alpha\\lvert \\uparrow \\rangle + \\beta\\lvert \\downarrow \\rangle$. Now, according to the Copenhagen interpretation, if we measure its spin along the $z$-axis, there is a probability $\\lvert \\alpha \\rvert ^{2}$ that the state collapses to $\\lvert \\uparrow \\rangle$ and a probability $\\lvert \\beta \\rvert ^{2}$ that it collapses to $\\lvert \\downarrow \\rangle$. Then, after a further $t$ seconds, the state will be $\\alpha\\lvert \\uparrow \\rangle + \\beta\\lvert \\downarrow \\rangle$ with probability $\\lvert \\alpha \\rvert ^{2}$ and $\\gamma\\lvert \\uparrow \\rangle + \\delta\\lvert \\downarrow \\rangle$ with probability $\\lvert \\beta \\rvert ^{2}$ for some $\\gamma$ and $\\delta$. Therefore, if we measure the particle's spin along the $z$-axis again, the probability that we will find it to be spin-up is $\\lvert \\alpha \\rvert ^{4}+\\lvert \\beta \\rvert ^{2}\\lvert \\gamma\\rvert ^{2}$.\nI am trying to understand the Everettian interpretation of quantum mechanics. According to the Everettian view as I understand it, after we make our first \"measurement\" in the scenario above, our measurement device and the particle would be entangled with a joint state of $\\alpha\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle + \\beta\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$, where $\\lvert\\frac{1}{2}\\rangle$ and $\\lvert-\\frac{1}{2}\\rangle$ are supposed to represent the states of the measuring device corresponding to it having measured spin-up and spin-down respectively. Before our second measurement then, our system would presumably be in the state $\\alpha^{2}\\lvert \\uparrow \\rangle \\lvert\\frac{1}{2}\\rangle + \\beta\\gamma\\lvert \\uparrow \\rangle\\lvert-\\frac{1}{2}\\rangle + \\alpha\\beta\\lvert \\downarrow \\rangle \\lvert\\frac{1}{2}\\rangle + \\beta\\delta\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$. Now, when we make our second measurement, the system is still supposed to evolve under the Schrodinger equation, which is linear, so to work out what will happen, we can just work out what state we would have ended up with if the state had initially been each of the four parts of our superposition and then take the appropriate linear combination. For example, if the initial state had been $\\beta\\gamma\\lvert \\uparrow \\rangle\\lvert-\\frac{1}{2}\\rangle$, then, we would presumably have ended up with the state $\\beta\\gamma\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle$, up to some phase, after we brought our measuring device and our particle back together. By this logic, it would seem that we would end up in the state $\\left(\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\right)\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle + \\left(e^{i\\theta_{2}}\\alpha\\beta + e^{i\\theta_{3}}\\beta\\delta\\right)\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$ for some $\\theta_{1}$, $\\theta_{2}$ and $\\theta_{3}$. The probability of measuring spin-up on our second measurement would apparently then be $\\lvert\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\rvert^{2}$.\nIf so, the predictions of the Copenhagen and Everettian interpretations of quantum mechanics could be in agreement with one another only if the Schrodinger equation made sure that $\\theta_{1}$ was such that $\\lvert \\alpha \\rvert ^{4}+\\lvert \\beta \\rvert ^{2}\\lvert \\gamma\\rvert ^{2} = \\lvert\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\rvert^{2}$, i.e. $\\Re\\left({\\alpha^{*}}^{2}e^{i\\theta_{1}}\\beta\\gamma\\right) = 0$. My questions then are:\n\nIs my above argument correct?\nIf so, does the Schrodinger equation make sure the phases are such that the predictions of the two interpretations agree and how can I see that it does so or does not?\n\n", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "b", "aspects": {"naturalness": "b", "cultural_appropriateness": "strong_b", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "smith", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:32:26", "duration_ms": 354345}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_11755::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_11755", "language": "French", "seed_prompt": "I came across Norton's dome and I don't agree that it proves anything.\nFirst, here's an obviously ridiculous and completely nonsensical example that I constructed from thinking about simple harmonic motion:\nConsider a particle of unit mass at rest in free space, at $x = 0$. Newton's second law states that $\\ddot{x} = 0$ at $t=0$. There's an obvious way to describe the particle's motion that satisfies Newton's second law of motion, and that is $x(t) = 0$. Let us denote this solution by $s_1(t)$.\nHowever, here's another non-obvious way to describe the particle's motion that satisfies Newton's second law:\n$$\nx(t) = \\begin{cases}\n\\sin(t - T) - t + T \\quad &\\text{if }\\ t \\ge T \\\\\n0 & \\text{if }\\ t < T\n\\end{cases}\n$$\nLet us denote this solution by $s_2(t)$.\nNotice that $s_2(t)$, $\\dot{s_2}(t)$ and $\\ddot{s_2}(t)$ all obviously equal $0$ for all $t < T$.\nNow consider time $t = T$. Notice that:\n$$\n\\begin{align*}\ns_2(T) & = \\sin(T - T) - T + T = 0 \\\\\n\\dot{s_2}(T) & = \\cos(T - T) - 1 = 0 \\\\\n\\ddot{s_2}(T) & = -\\sin(T - T) = 0\n\\end{align*}\n$$\nThus, $s_2(t)$ is a completely valid solution which obeys the constraint that the particle is at rest at time $t=0$. However, notice that as $t$ grows beyond $T$, the particle starts to spontaneously shoot off to infinity in the negative $x$ direction. Also notice that $T$ is an arbitrarily chosen constant, it could assume any positive real number as its value.\nThus, the second solution states that the particle could spontaneously start shooting off to infinity in the negative $x$ direction after some arbitrary amount of time $T$.\nSo we have proven that Newtonian mechanics isn't deterministic, can harbor uncaused events, yada yada.\nWhy are Norton's dome and my example not valid?\nThey're incorrect because in those solutions, the particle isn't really at rest when the spontaneous motion starts occurring. Notice that the fourth derivative of $r(t) = \\frac{d^4}{dt^4} \\left( \\frac{(t - T)^4}{144} \\right) = \\frac{1}{6}$, and that the third derivative of $s_2(t)$ at $t=T$ is equal to $-\\cos(T-T) = -1$.\nClearly Norton believes that an object being \"at rest\" does not necessitate that $\\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n \\ge 2 $. He believes that Newton's first law does not imply $ \\frac{d^2 \\vec{p}}{dt^2} \\implies \\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n \\ge 2 $.\nHowever, believing this leads to all sorts of completely nonsensical predictions for even the simplest system one can think of, a particle at rest in free space, as I have shown above.\nTherefore, it only makes sense to believe that $ \\frac{d^2 \\vec{p}}{dt^2} = \\vec{0} \\implies \\frac{d^n \\vec{p}}{dt^n} = \\vec{0} \\ \\forall \\ n > 2$.\nSo, coming to my questions: Are my arguments valid? Is Norton's second solution just plain incorrect? If you disagree with my arguments and instead agree with Norton, what do you think of my completely nonsensical second solution for a particle at rest in free space?\n", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "strong_b", "aspects": {"naturalness": "strong_b", "cultural_appropriateness": "tie_both_bad", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "smith", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:23:46", "duration_ms": 290119}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_40351::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_40351", "language": "French", "seed_prompt": "Benzenediazonium tetrafluoroborate decomposes in dimethyl sulfoxide with evolution of nitrogen and phenylation of any aromatic present. The following isomer ratios were obtained for the phenylation of toluene, chlorobenzene, and nitrobenzene.\\n\\n| Compound | \\( \\text{p/o} \\) | \\( \\text{m/l} \\) | \\( \\text{o/m} \\) |\\n| -------- | ---------------- | ---------------- | ---------------- |\\n| PhCH\\(_3\\) | 47:3 | 2:1·1 | 31·6 |\\n| PhCl | 48 | 2·40 | 27·8 |\\n| PhNO\\(_2\\) | 38·1 | 48·4 | 13·5 |\\n\\n## Miscellaneous Problems\\n\\nThe rates of reaction correlate with \\(\\sigma^*\\) and the value of \\(\\rho\\) is −0.74. The isomer ratios are the same for anisole-d\\(_6\\) and anisole. Is phenylunion under these conditions free radical or electrophilic and what is the rate-determining step?", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "strong_a", "aspects": {"naturalness": "strong_a", "cultural_appropriateness": "tie_both_bad", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "teacher", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:26:32", "duration_ms": 166051}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_42290::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_42290", "language": "French", "seed_prompt": "Suppose we have a spin $\\frac{1}{2}$ particle in the spin-up state along the $z$-axis, $\\lvert \\uparrow \\rangle$, and after $t$ seconds of evolution under the Schrodinger equation it is in state $\\alpha\\lvert \\uparrow \\rangle + \\beta\\lvert \\downarrow \\rangle$. Now, according to the Copenhagen interpretation, if we measure its spin along the $z$-axis, there is a probability $\\lvert \\alpha \\rvert ^{2}$ that the state collapses to $\\lvert \\uparrow \\rangle$ and a probability $\\lvert \\beta \\rvert ^{2}$ that it collapses to $\\lvert \\downarrow \\rangle$. Then, after a further $t$ seconds, the state will be $\\alpha\\lvert \\uparrow \\rangle + \\beta\\lvert \\downarrow \\rangle$ with probability $\\lvert \\alpha \\rvert ^{2}$ and $\\gamma\\lvert \\uparrow \\rangle + \\delta\\lvert \\downarrow \\rangle$ with probability $\\lvert \\beta \\rvert ^{2}$ for some $\\gamma$ and $\\delta$. Therefore, if we measure the particle's spin along the $z$-axis again, the probability that we will find it to be spin-up is $\\lvert \\alpha \\rvert ^{4}+\\lvert \\beta \\rvert ^{2}\\lvert \\gamma\\rvert ^{2}$.\nI am trying to understand the Everettian interpretation of quantum mechanics. According to the Everettian view as I understand it, after we make our first \"measurement\" in the scenario above, our measurement device and the particle would be entangled with a joint state of $\\alpha\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle + \\beta\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$, where $\\lvert\\frac{1}{2}\\rangle$ and $\\lvert-\\frac{1}{2}\\rangle$ are supposed to represent the states of the measuring device corresponding to it having measured spin-up and spin-down respectively. Before our second measurement then, our system would presumably be in the state $\\alpha^{2}\\lvert \\uparrow \\rangle \\lvert\\frac{1}{2}\\rangle + \\beta\\gamma\\lvert \\uparrow \\rangle\\lvert-\\frac{1}{2}\\rangle + \\alpha\\beta\\lvert \\downarrow \\rangle \\lvert\\frac{1}{2}\\rangle + \\beta\\delta\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$. Now, when we make our second measurement, the system is still supposed to evolve under the Schrodinger equation, which is linear, so to work out what will happen, we can just work out what state we would have ended up with if the state had initially been each of the four parts of our superposition and then take the appropriate linear combination. For example, if the initial state had been $\\beta\\gamma\\lvert \\uparrow \\rangle\\lvert-\\frac{1}{2}\\rangle$, then, we would presumably have ended up with the state $\\beta\\gamma\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle$, up to some phase, after we brought our measuring device and our particle back together. By this logic, it would seem that we would end up in the state $\\left(\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\right)\\lvert \\uparrow \\rangle\\lvert\\frac{1}{2}\\rangle + \\left(e^{i\\theta_{2}}\\alpha\\beta + e^{i\\theta_{3}}\\beta\\delta\\right)\\lvert \\downarrow \\rangle\\lvert-\\frac{1}{2}\\rangle$ for some $\\theta_{1}$, $\\theta_{2}$ and $\\theta_{3}$. The probability of measuring spin-up on our second measurement would apparently then be $\\lvert\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\rvert^{2}$.\nIf so, the predictions of the Copenhagen and Everettian interpretations of quantum mechanics could be in agreement with one another only if the Schrodinger equation made sure that $\\theta_{1}$ was such that $\\lvert \\alpha \\rvert ^{4}+\\lvert \\beta \\rvert ^{2}\\lvert \\gamma\\rvert ^{2} = \\lvert\\alpha^{2} + e^{i\\theta_{1}}\\beta\\gamma\\rvert^{2}$, i.e. $\\Re\\left({\\alpha^{*}}^{2}e^{i\\theta_{1}}\\beta\\gamma\\right) = 0$. My questions then are:\n\nIs my above argument correct?\nIf so, does the Schrodinger equation make sure the phases are such that the predictions of the two interpretations agree and how can I see that it does so or does not?\n\n", "pair": "smith_vs_teacher", "system_left": "teacher", "system_right": "smith", "overall": "b", "aspects": {"naturalness": "b", "cultural_appropriateness": "strong_b", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "smith", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:32:26", "duration_ms": 354345}
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{"comparison_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_43332::smith_vs_teacher", "source_id": "allenai/OpenThoughts3-full-filtered-science-no-cot-decontam-v2_tmp_ids_43332", "language": "French", "seed_prompt": "Consider the hypothetical cyclic phosphonitrilic compound \\([N_4(\\text{PX}_2)_2]\\). After taking care of the \\(\\sigma\\) bonding, it is assumed that the participating atomic orbitals for the \\(\\pi\\) bonds around the ring are the four nitrogen \\(p_\\pi\\) orbitals and two phosphorus \\(d_\\pi\\) orbitals. These orbitals are shown and labeled below, where only the signed lobes above the molecular plane are displayed.\\n\\n#### (i)\\nTaking advantage of the \\(C_{2v}\\) symmetry of the system, display the six \\(\\pi\\) molecular orbitals formed by the aforementioned atomic orbitals pictorially and label the orbitals with their symmetry. Also, arrange the molecular orbitals in order of increasing energy. Rationalize your ordering clearly.\\n\\n#### (ii)\\nWrite down the ground electronic configuration and ground state of this system with six π electrons.\\n\\n#### (iii)\\nWhat is the lowest-energy symmetry-allowed electronic transition for this system? Clearly write out the states as well as the configurations involved.", "pair": "smith_vs_teacher", "system_left": "smith", "system_right": "teacher", "overall": "a", "aspects": {"naturalness": "a", "cultural_appropriateness": "tie_both_bad", "instruction_following": "tie_both_good", "coherence": "tie_both_good"}, "comment": "", "instance_feedback": "", "winner_system": "smith", "annotator_id": "happy_bardeen", "session_id": "ee2c9108e413443bb528f88138d6f34c", "annotated_at": "2026-09-25T09:35:05", "duration_ms": 158125}
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