{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0001","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdc171ca22d8d66f3c5ea8adaaed9d0bd1e0a35866c1526b58c6ed7bbeca8383","family_id":"qf-proof-modus-ponens","seed":20260922,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 可被 4 整除\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 是偶數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is divisible by 4” and Q mean “the integer n is even”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0002","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd91db8655e65042dfae0fb9ec291db28b61bdf3163e319fededb1beaa2e2add","family_id":"qf-proof-modus-ponens","seed":20260923,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 可被 6 整除\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 可被 3 整除\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is divisible by 6” and Q mean “the integer n is divisible by 3”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0003","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c045d6e8bfb1bfcd2b79114c788ddf781daaf54632594307d7866b4774e75044","family_id":"qf-proof-modus-ponens","seed":20260924,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是 9 的倍數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 是 3 的倍數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is a multiple of 9” and Q mean “the integer n is a multiple of 3”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0004","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ea253532b4c3bd34677dea7ea268898323ba0b01748fa860b1acf184358b990","family_id":"qf-proof-modus-ponens","seed":20260925,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是質數且大於 2\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 是奇數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is prime and greater than 2” and Q mean “the integer n is odd”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0005","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5311abfaec9f565c3eee67cdfc2334a14dae768bbb3a3edf143eb4ca79d23935","family_id":"qf-proof-modus-ponens","seed":20260926,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"三角形有三條相等的邊\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"三角形至少有兩條相等的邊\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the triangle has three equal sides” and Q mean “the triangle has at least two equal sides”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0006","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63007a0349d7e25512adc279077a21679b182c6c96c26e00721ecbb46de3839f","family_id":"qf-proof-modus-ponens","seed":20260927,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"四邊形是正方形\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"四邊形有四個直角\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the quadrilateral is a square” and Q mean “the quadrilateral has four right angles”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0007","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"444880acdaf905eba922c4005607c67f38639bbe4d7e6f2f780baa32ee81191b","family_id":"qf-proof-modus-ponens","seed":20260928,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是 10 的倍數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 是 5 的倍數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is a multiple of 10” and Q mean “the integer n is a multiple of 5”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0008","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0c35df4aa568cf5d65bc22f9958b40833b04c1b2c03c25873ea0243a98ded5b","family_id":"qf-proof-modus-ponens","seed":20260929,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是 12 的倍數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 是 4 的倍數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is a multiple of 12” and Q mean “the integer n is a multiple of 4”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0009","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e195ca452eb79b0fe74b6569832e0dd8cde01f028e6dc81c1b5704e069c1dfcb","family_id":"qf-proof-modus-ponens","seed":20260930,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是完全平方數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"存在整數 k 使 n=k²\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is a perfect square” and Q mean “there exists an integer k such that n=k²”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0010","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc498126b7491919dfdb8251f0a8e72af68c807389c70f9f5a80c27ac20c94f6","family_id":"qf-proof-modus-ponens","seed":20260931,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"有理數 r 等於 0\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"r 的絕對值等於 0\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the rational number r equals 0” and Q mean “the absolute value of r equals 0”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0011","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8457a053d6a1100daeaef151efdec2b560729a8139f389058275a5ec98f3fbbb","family_id":"qf-proof-modus-ponens","seed":20260932,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"集合 A 是空集合\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"A 沒有元素\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “set A is empty” and Q mean “A has no elements”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0012","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb56b4a07972606e6fbac0eaacefe26d6e908d31a81d96a247d7398af55b7a6a","family_id":"qf-proof-modus-ponens","seed":20260933,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"函數 f 是常數函數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"任兩個輸入有相同輸出\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “function f is constant” and Q mean “any two inputs have the same output”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0013","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4cd7c9bda092437bf84d03d45ced1e0f0c7ae57247c4bad174f1af968450acb","family_id":"qf-proof-modus-ponens","seed":20260934,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"角的量值是 90°\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"角是直角\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the angle measure is 90°” and Q mean “the angle is a right angle”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0014","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8bd1c6e6809f5eab56a8890058ecff80b4d4430b0fa681b0354464e5cb31ad6","family_id":"qf-proof-modus-ponens","seed":20260935,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"角的量值是 180°\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"角是平角\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the angle measure is 180°” and Q mean “the angle is a straight angle”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0015","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39487f47a20c830f541707983b998db356e74943ed00a16cf22436dcb6f37bf6","family_id":"qf-proof-modus-ponens","seed":20260936,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 可被 8 整除\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 可被 2 整除\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is divisible by 8” and Q mean “the integer n is divisible by 2”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0016","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a979144d3ed1885819e32f4fa7096ed267de6b4cf2175e515c2f3b450e978fc4","family_id":"qf-proof-modus-ponens","seed":20260937,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是 15 的倍數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"整數 n 可被 5 整除\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is a multiple of 15” and Q mean “the integer n is divisible by 5”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0017","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cca031592d4f26a85d33eebd27b43732bdb7486462c06a093200c4f860199471","family_id":"qf-proof-modus-ponens","seed":20260938,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"數 x 大於 7\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"x 大於 0\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the number x is greater than 7” and Q mean “x is greater than 0”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0018","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbae3f1c5aa2b33fe9f687b54f3a25c1993b8cb37018fee8fc4b388957cf88b3","family_id":"qf-proof-modus-ponens","seed":20260939,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"數 x 小於 -3\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"x 小於 0\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the number x is less than -3” and Q mean “x is less than 0”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0019","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36eaac9f4b8c4b10dd27e68144efeafe46bcc943af42556ebea117ed95d2eea7","family_id":"qf-proof-modus-ponens","seed":20260940,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"多邊形是五邊形\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"多邊形有五條邊\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the polygon is a pentagon” and Q mean “the polygon has five sides”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0020","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a071651fbe7b668ff66925c6d0b30fbf7b2520c87f910fa3c3276b314b9cb2f","family_id":"qf-proof-modus-ponens","seed":20260941,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"矩形的長與寬相等\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"矩形是正方形\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the rectangle has equal length and width” and Q mean “the rectangle is a square”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0021","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efb691a44c6b35686626ab7c6b42a3d18221f9f2494714efb6756d9fbbe31f6d","family_id":"qf-proof-modus-ponens","seed":20260942,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 是偶數\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"n² 是偶數\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is even” and Q mean “n² is even”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0022","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8dadbd29f9dfc6e6ba9e894b9adbb588589cb55295195cd3c6ef6f5ed46f9a8c","family_id":"qf-proof-modus-ponens","seed":20260943,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"整數 n 可被 3 整除\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"n² 可被 3 整除\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the integer n is divisible by 3” and Q mean “n² is divisible by 3”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0023","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d681a8d9274c4cf2a7ad6d5f7b94816e18e91de591bfd7c8cc9fb24f74dacda8","family_id":"qf-proof-modus-ponens","seed":20260944,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"分數 a/b 的分母 b 不等於 0\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"a/b 是有定義的分數表示\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “the denominator b of a/b is nonzero” and Q mean “a/b is a defined fractional representation”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0024","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80ce8fa8fa72cf6d9a0bef6e58c4338ee827b9efba75a4a1da9671976d2c0f07","family_id":"qf-proof-modus-ponens","seed":20260945,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"集合 A 是 B 的子集\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"A 中每個元素也在 B 中\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “set A is a subset of B” and Q mean “every element of A is also in B”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-ponens-0025","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58596fae2c657f109acfc6363cbae4dc57ec837e1ae153e1651f83276afbec2d","family_id":"qf-proof-modus-ponens","seed":20260946,"semantic_template":"qf-proof-modus-ponens/{\"conclusion\":\"Q\",\"inference_rule\":\"modus_ponens\",\"p\":\"命題 R 與 S 都成立\",\"premises\":[\"P→Q\",\"P\"],\"q\":\"命題 R 成立\"}"},"fields":{"title":"Modus Ponens","stem":"Let P mean “propositions R and S both hold” and Q mean “proposition R holds”. The stated premises are P→Q and P, and the proposed conclusion is Q. Under modus ponens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and P, exactly the form of modus ponens, so Q follows."},"skills":["material implication","modus ponens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0001","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5d9315f8bf13e44a90a7102a10b8d443052d1dd6efb590115605be3b8788d0f","family_id":"qf-proof-modus-tollens","seed":20261022,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 可被 4 整除\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 是偶數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is divisible by 4” and Q mean “the integer n is even”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0002","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77642ac27da91daaa0f80c081535a8266be47987705195833ac7087dcb604094","family_id":"qf-proof-modus-tollens","seed":20261023,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 可被 6 整除\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 可被 3 整除\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is divisible by 6” and Q mean “the integer n is divisible by 3”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0003","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a67598bcfd1fe3c2174214b4bb1efbcb92a43be55f0f744a568180e0e3ac8508","family_id":"qf-proof-modus-tollens","seed":20261024,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是 9 的倍數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 是 3 的倍數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is a multiple of 9” and Q mean “the integer n is a multiple of 3”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0004","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b41f49007e969fc118fb6f453c515e5e569b3dca0fd80a567ea3cf3bde8373c","family_id":"qf-proof-modus-tollens","seed":20261025,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是質數且大於 2\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 是奇數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is prime and greater than 2” and Q mean “the integer n is odd”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0005","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40dc152e35e0ab811e5d94d6ed1a34876ddc11ef6e0fdfd72a70be56f3883d76","family_id":"qf-proof-modus-tollens","seed":20261026,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"三角形有三條相等的邊\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"三角形至少有兩條相等的邊\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the triangle has three equal sides” and Q mean “the triangle has at least two equal sides”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0006","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8de3855810caeef87bc00ae6daa6544cc21304e28bbd1749ec7127081a65e9e5","family_id":"qf-proof-modus-tollens","seed":20261027,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"四邊形是正方形\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"四邊形有四個直角\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the quadrilateral is a square” and Q mean “the quadrilateral has four right angles”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0007","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8da2a589c164b5ebb9d17fe8d4429ef90cfedca63410d294053c0ea3bbe1453","family_id":"qf-proof-modus-tollens","seed":20261028,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是 10 的倍數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 是 5 的倍數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is a multiple of 10” and Q mean “the integer n is a multiple of 5”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0008","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9109e803a2157d379d8f58a9a28ee1c821508cf2b2b12082af8e5c6332786f6","family_id":"qf-proof-modus-tollens","seed":20261029,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是 12 的倍數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 是 4 的倍數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is a multiple of 12” and Q mean “the integer n is a multiple of 4”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0009","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce5d140a712667771f31d96a1ab5db58587aef2e999e8cc88a1ff56df7ee309a","family_id":"qf-proof-modus-tollens","seed":20261030,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是完全平方數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"存在整數 k 使 n=k²\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is a perfect square” and Q mean “there exists an integer k such that n=k²”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0010","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4fca13524bc78ac63283659d3102444a0356964082562bdb90bddfa0e98e11e0","family_id":"qf-proof-modus-tollens","seed":20261031,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"有理數 r 等於 0\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"r 的絕對值等於 0\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the rational number r equals 0” and Q mean “the absolute value of r equals 0”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0011","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"feb5ec34a741f63b111045d6186b2f50de6e627cacb66f600abe34d5e6bc5960","family_id":"qf-proof-modus-tollens","seed":20261032,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"集合 A 是空集合\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"A 沒有元素\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “set A is empty” and Q mean “A has no elements”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0012","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb222fd539a50cff038908618738c313337b5c0e2080896c2a89db70d25910ca","family_id":"qf-proof-modus-tollens","seed":20261033,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"函數 f 是常數函數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"任兩個輸入有相同輸出\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “function f is constant” and Q mean “any two inputs have the same output”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0013","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9997e6938bda286ac2d237ab1b979b94e612b8f69b3567cb88b63776e3fa1cff","family_id":"qf-proof-modus-tollens","seed":20261034,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"角的量值是 90°\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"角是直角\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the angle measure is 90°” and Q mean “the angle is a right angle”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0014","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e86cdbcbe262e4197d41fbd9ae09e4fcbc7c94d0a640d9fb3c273c901394147f","family_id":"qf-proof-modus-tollens","seed":20261035,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"角的量值是 180°\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"角是平角\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the angle measure is 180°” and Q mean “the angle is a straight angle”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0015","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e5e93fa76ab6f23026ae0865b3b37f7d26b0f44a6968e292db31b02ef36402f","family_id":"qf-proof-modus-tollens","seed":20261036,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 可被 8 整除\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 可被 2 整除\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is divisible by 8” and Q mean “the integer n is divisible by 2”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0016","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7744e26156450f9cd44074ec967d68fd1f3bd09105a8af39496d802adaea81f2","family_id":"qf-proof-modus-tollens","seed":20261037,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是 15 的倍數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"整數 n 可被 5 整除\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is a multiple of 15” and Q mean “the integer n is divisible by 5”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0017","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4f17461eaf5102583a9a40eb623903702589f62d07bf8bc2acb158d7fc61582","family_id":"qf-proof-modus-tollens","seed":20261038,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"數 x 大於 7\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"x 大於 0\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the number x is greater than 7” and Q mean “x is greater than 0”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0018","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba2c88d078f2273339740077e47e8587cd54dd641468c11135310efb14e31061","family_id":"qf-proof-modus-tollens","seed":20261039,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"數 x 小於 -3\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"x 小於 0\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the number x is less than -3” and Q mean “x is less than 0”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0019","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"707b6ecfe9a745c23788689ae66053a13ef83426bbaa91b53464122eee296cb3","family_id":"qf-proof-modus-tollens","seed":20261040,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"多邊形是五邊形\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"多邊形有五條邊\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the polygon is a pentagon” and Q mean “the polygon has five sides”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0020","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a379ce74b79b4b61f772ea159d9d17dd81dc694a440ccb0b9a4f422e03fe6a74","family_id":"qf-proof-modus-tollens","seed":20261041,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"矩形的長與寬相等\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"矩形是正方形\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the rectangle has equal length and width” and Q mean “the rectangle is a square”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0021","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63485847c77e6830e4916de613cd01990733c81cca0e3046fc596675f4509a22","family_id":"qf-proof-modus-tollens","seed":20261042,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 是偶數\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"n² 是偶數\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is even” and Q mean “n² is even”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0022","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da5413374566f90f67768a9dce08c27f4cede0161b18047c3b714d732cc9a3fb","family_id":"qf-proof-modus-tollens","seed":20261043,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"整數 n 可被 3 整除\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"n² 可被 3 整除\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the integer n is divisible by 3” and Q mean “n² is divisible by 3”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0023","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7877b750d82f1af347b3f9e02345c2ac8cf9dc6ac74b8fa0bfee86211a3d325","family_id":"qf-proof-modus-tollens","seed":20261044,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"分數 a/b 的分母 b 不等於 0\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"a/b 是有定義的分數表示\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “the denominator b of a/b is nonzero” and Q mean “a/b is a defined fractional representation”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0024","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0759c777b8f95aa66511d14d9901a62767a52cf53cdcab45c45055593d5a68a8","family_id":"qf-proof-modus-tollens","seed":20261045,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"集合 A 是 B 的子集\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"A 中每個元素也在 B 中\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “set A is a subset of B” and Q mean “every element of A is also in B”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-modus-tollens-0025","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"074f93eed17845bb2401d5b44e4a8ebda71426ac4f1b21e709231a0bea7e3317","family_id":"qf-proof-modus-tollens","seed":20261046,"semantic_template":"qf-proof-modus-tollens/{\"conclusion\":\"¬P\",\"inference_rule\":\"modus_tollens\",\"p\":\"命題 R 與 S 都成立\",\"premises\":[\"P→Q\",\"¬Q\"],\"q\":\"命題 R 成立\"}"},"fields":{"title":"Modus Tollens","stem":"Let P mean “propositions R and S both hold” and Q mean “proposition R holds”. The stated premises are P→Q and ¬Q, and the proposed conclusion is ¬P. Under modus tollens, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Can","choices":["Can","Cannot"],"explanation":"The premises are P→Q and ¬Q, exactly the form of modus tollens, so ¬P follows."},"skills":["material implication","modus tollens"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0001","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e324f82a8c863674cc62f46705557120af9f03f2d799cbf06bea1e5f6d7dbd15","family_id":"qf-proof-invalid-converse","seed":20261122,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 可被 4 整除\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 是偶數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is divisible by 4” and Q mean “the integer n is even”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0002","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e195ff9b97248204baf9271e7e4de958114f919395986c30b74550147ba3f4f","family_id":"qf-proof-invalid-converse","seed":20261123,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 可被 6 整除\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 可被 3 整除\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is divisible by 6” and Q mean “the integer n is divisible by 3”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0003","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d81e0efe0ccd300861f06f1b7528967550b71d0b904f4d208f3af2488a82adbf","family_id":"qf-proof-invalid-converse","seed":20261124,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是 9 的倍數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 是 3 的倍數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is a multiple of 9” and Q mean “the integer n is a multiple of 3”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0004","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2b4a2b888efae03ff512c406594405743419744609e82cf790ee3a89784c52c","family_id":"qf-proof-invalid-converse","seed":20261125,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是質數且大於 2\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 是奇數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is prime and greater than 2” and Q mean “the integer n is odd”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0005","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70d902e628996f82b77b84502ea81271a4fe76c9324c782fa513eb54ef6ab339","family_id":"qf-proof-invalid-converse","seed":20261126,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"三角形有三條相等的邊\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"三角形至少有兩條相等的邊\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the triangle has three equal sides” and Q mean “the triangle has at least two equal sides”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0006","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7c7d29a66a864c6a9c6bc2b30c2ce64917568f761c099293f2cdf547ab6ae41","family_id":"qf-proof-invalid-converse","seed":20261127,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"四邊形是正方形\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"四邊形有四個直角\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the quadrilateral is a square” and Q mean “the quadrilateral has four right angles”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0007","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7adfdc767bb70315c2679797f0e82bc6c647b5f21923b00e7f4c70e8455711eb","family_id":"qf-proof-invalid-converse","seed":20261128,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是 10 的倍數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 是 5 的倍數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is a multiple of 10” and Q mean “the integer n is a multiple of 5”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0008","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"089c89618a8a88c1a5ba38040e900c7f21f8cc5486ef9cc13a998c155421ff22","family_id":"qf-proof-invalid-converse","seed":20261129,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是 12 的倍數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 是 4 的倍數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is a multiple of 12” and Q mean “the integer n is a multiple of 4”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0009","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15b227e81844d30141ce94494dae153f42b3a4b9b84ce908310f5ad4cc6caf02","family_id":"qf-proof-invalid-converse","seed":20261130,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是完全平方數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"存在整數 k 使 n=k²\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is a perfect square” and Q mean “there exists an integer k such that n=k²”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0010","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"376e76accf077cd118de62cafb553b77471683452793858f5bdcf5124eea2480","family_id":"qf-proof-invalid-converse","seed":20261131,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"有理數 r 等於 0\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"r 的絕對值等於 0\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the rational number r equals 0” and Q mean “the absolute value of r equals 0”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0011","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70d4c75914bb930fc277b831417ad8751664f1e5d6cffb50ad2f4528c4035767","family_id":"qf-proof-invalid-converse","seed":20261132,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"集合 A 是空集合\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"A 沒有元素\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “set A is empty” and Q mean “A has no elements”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0012","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a30c72f6ab3149fb734747185373f632cb87605d1fd95c12a6df2b539c96a42","family_id":"qf-proof-invalid-converse","seed":20261133,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"函數 f 是常數函數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"任兩個輸入有相同輸出\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “function f is constant” and Q mean “any two inputs have the same output”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0013","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fda5e8d93fb0858bf1ed2fdaa898d28644b1858f7f721474448d2412c61549aa","family_id":"qf-proof-invalid-converse","seed":20261134,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"角的量值是 90°\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"角是直角\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the angle measure is 90°” and Q mean “the angle is a right angle”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0014","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39a8ad054fedce55d6af2de5f4e548fbd8f0595cccff81062712f6edb7a2d7e9","family_id":"qf-proof-invalid-converse","seed":20261135,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"角的量值是 180°\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"角是平角\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the angle measure is 180°” and Q mean “the angle is a straight angle”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0015","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cecdba8182ba79f9365684afc1ae48bbd6dffbdfd34ea4daf576871ef7d771a4","family_id":"qf-proof-invalid-converse","seed":20261136,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 可被 8 整除\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 可被 2 整除\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is divisible by 8” and Q mean “the integer n is divisible by 2”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0016","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50ca6d9b389ca0c356e062ca839d71bcddd6549074501a92450e60dfa75e4809","family_id":"qf-proof-invalid-converse","seed":20261137,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是 15 的倍數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"整數 n 可被 5 整除\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is a multiple of 15” and Q mean “the integer n is divisible by 5”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0017","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27df4e974da2c4d46ea8922e797cb94bd5dfec97ccec0b30c689dadf69abfa2d","family_id":"qf-proof-invalid-converse","seed":20261138,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"數 x 大於 7\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"x 大於 0\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the number x is greater than 7” and Q mean “x is greater than 0”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0018","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b757cbfacd6555887fe41778105c75faec05167281b2951436b88b2b00366702","family_id":"qf-proof-invalid-converse","seed":20261139,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"數 x 小於 -3\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"x 小於 0\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the number x is less than -3” and Q mean “x is less than 0”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0019","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31c35490556e364970f32069efe2d9ca948ec6d785218edcb0ad101ccfa6c97d","family_id":"qf-proof-invalid-converse","seed":20261140,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"多邊形是五邊形\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"多邊形有五條邊\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the polygon is a pentagon” and Q mean “the polygon has five sides”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0020","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf9f9be1c2b69baf830c04a9806464f18f46dd26b8fc52b17df48bb5d57992a9","family_id":"qf-proof-invalid-converse","seed":20261141,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"矩形的長與寬相等\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"矩形是正方形\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the rectangle has equal length and width” and Q mean “the rectangle is a square”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0021","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b662188acff4f69e220523143339c7e16fe6c03fa32fb0ee6d4eab14a7f8735","family_id":"qf-proof-invalid-converse","seed":20261142,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 是偶數\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"n² 是偶數\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is even” and Q mean “n² is even”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0022","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17db1d039a1fb73174b613b414b7ce30cbe86614b1b99ae2b134c92a6e3f5b47","family_id":"qf-proof-invalid-converse","seed":20261143,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"整數 n 可被 3 整除\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"n² 可被 3 整除\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the integer n is divisible by 3” and Q mean “n² is divisible by 3”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0023","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c71d07b02046e351567e63bdb5be966904916e719ed892d34e9ebf258619193","family_id":"qf-proof-invalid-converse","seed":20261144,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"分數 a/b 的分母 b 不等於 0\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"a/b 是有定義的分數表示\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “the denominator b of a/b is nonzero” and Q mean “a/b is a defined fractional representation”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0024","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50cf1ea95006b6589219c6937704e079061932a3723c691f98eee18c7981b4a1","family_id":"qf-proof-invalid-converse","seed":20261145,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"集合 A 是 B 的子集\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"A 中每個元素也在 B 中\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “set A is a subset of B” and Q mean “every element of A is also in B”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-invalid-converse-0025","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a275cf64a93b357ff977eba7d087045f6e1ae54556acfdb88fc5eced22f0161","family_id":"qf-proof-invalid-converse","seed":20261146,"semantic_template":"qf-proof-invalid-converse/{\"conclusion\":\"P\",\"counterexample_truth_assignment\":{\"P\":false,\"Q\":true},\"inference_rule\":\"invalid_converse\",\"p\":\"命題 R 與 S 都成立\",\"premises\":[\"P→Q\",\"Q\"],\"q\":\"命題 R 成立\"}"},"fields":{"title":"Invalid Affirmation of the Consequent","stem":"Let P mean “propositions R and S both hold” and Q mean “proposition R holds”. The stated premises are P→Q and Q, and the proposed conclusion is P. Under the converse inference, can the conclusion be derived in general?","answer":{"kind":"choice","value":"Cannot","choices":["Can","Cannot"],"explanation":"Take P=false and Q=true. Then P→Q and Q are both true while P is false, so P does not follow in general."},"skills":["material implication","invalid converse","counterexample truth assignment"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0001","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a21bc36e30e3ee2c08724df58daa761e254d318f4d4518c7e90a08ba58acd86","family_id":"qf-proof-counterexample-universal","seed":20261222,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"even\",\"universal_statement\":\"每個整數都是偶數\",\"witness\":1}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is even.” and the stated witness n=1. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=1 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0002","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ecd98f51551f25e14738ec5d43bea71f2c346523fed893cbfa57dc39c49ffee8","family_id":"qf-proof-counterexample-universal","seed":20261223,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"positive\",\"universal_statement\":\"每個整數都大於 0\",\"witness\":0}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than 0.” and the stated witness n=0. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=0 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0003","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83e7be74192b8ec8f5d98d095ccbd97abfdd10ff3f927cc0e42a673dce2e8814","family_id":"qf-proof-counterexample-universal","seed":20261224,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"nonnegative\",\"universal_statement\":\"每個整數都大於或等於 0\",\"witness\":-1}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than or equal to 0.” and the stated witness n=-1. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-1 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0004","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe7cf1a1b68c043cefc36eb7aae4fd49d773ceb9b667a464a52a0f27300da266","family_id":"qf-proof-counterexample-universal","seed":20261225,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":3,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 3 的倍數\",\"witness\":1}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 3.” and the stated witness n=1. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=1 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0005","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"202afca52e278d50a7bc5731a38d0c2317708ec7cd17ea68694e4053f12d6d25","family_id":"qf-proof-counterexample-universal","seed":20261226,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":5,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 5 的倍數\",\"witness\":2}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 5.” and the stated witness n=2. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=2 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0006","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6a2e11c926b7e9fd4a836c144ce1a541e9c56c6d1f6d02e93d73b03228328a4","family_id":"qf-proof-counterexample-universal","seed":20261227,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"zero\",\"universal_statement\":\"每個整數都等於 0\",\"witness\":1}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer equals 0.” and the stated witness n=1. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=1 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0007","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"809771c101ea47bc0425898eac0ba79b0c0474562286c53c0a367eca692733f2","family_id":"qf-proof-counterexample-universal","seed":20261228,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":5,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"greater\",\"universal_statement\":\"每個整數都大於 5\",\"witness\":5}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than 5.” and the stated witness n=5. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=5 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0008","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"694b3556ccd162ee80dd17c23602685389f6830d1d55c09902f0503777407a4c","family_id":"qf-proof-counterexample-universal","seed":20261229,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":10,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"less\",\"universal_statement\":\"每個整數都小於 10\",\"witness\":10}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is less than 10.” and the stated witness n=10. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=10 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0009","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a88a0874f01795b6f64ccb870f98530e6ec7db1076fdfda78eea10cf086af17","family_id":"qf-proof-counterexample-universal","seed":20261230,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"even\",\"universal_statement\":\"每個整數都是偶數\",\"witness\":-3}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is even.” and the stated witness n=-3. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-3 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0010","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b545d532f49312de6990b16887fff59fbf928781f736e780be354c5ca31e8df3","family_id":"qf-proof-counterexample-universal","seed":20261231,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"positive\",\"universal_statement\":\"每個整數都大於 0\",\"witness\":-1}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than 0.” and the stated witness n=-1. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-1 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0011","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d36e4b9edee27b5201fc81fb6d160522a2a0621ba8c64739fef9ce274cd0e049","family_id":"qf-proof-counterexample-universal","seed":20261232,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":7,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 7 的倍數\",\"witness\":3}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 7.” and the stated witness n=3. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=3 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0012","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14e7d6f3ea0808546a19f9f48e2fffcb35bc229c19058b2704bf92b24ddcc94f","family_id":"qf-proof-counterexample-universal","seed":20261233,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"zero\",\"universal_statement\":\"每個整數都等於 0\",\"witness\":-2}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer equals 0.” and the stated witness n=-2. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-2 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0013","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa289f8f080dfbeed7ad21a576d1be26c97250d32d9b1047a8d70292267f2a12","family_id":"qf-proof-counterexample-universal","seed":20261234,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":-4,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"greater\",\"universal_statement\":\"每個整數都大於 -4\",\"witness\":-4}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than -4.” and the stated witness n=-4. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-4 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0014","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3308c1199557a7f80213cd59fa30d97a4bf2c14e8c15ce2eeb3a6acbb549fb26","family_id":"qf-proof-counterexample-universal","seed":20261235,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":0,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"less\",\"universal_statement\":\"每個整數都小於 0\",\"witness\":0}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is less than 0.” and the stated witness n=0. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=0 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0015","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f5ac3f38bb83483a73e96243cc0d6689d420d03690006061864642d634a7aa9","family_id":"qf-proof-counterexample-universal","seed":20261236,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":2,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 2 的倍數\",\"witness\":5}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 2.” and the stated witness n=5. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=5 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0016","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d855d2dd6d8d7c2999c04f8f2219fedd40c206770b4c481d40011f4d2ab7875","family_id":"qf-proof-counterexample-universal","seed":20261237,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"nonnegative\",\"universal_statement\":\"每個整數都大於或等於 0\",\"witness\":-9}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than or equal to 0.” and the stated witness n=-9. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-9 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0017","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5c1662fb4c76c2cb791038b6d2f51c08949212c1eb5d7d2b3bad78996080d97","family_id":"qf-proof-counterexample-universal","seed":20261238,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":11,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 11 的倍數\",\"witness\":4}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 11.” and the stated witness n=4. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=4 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0018","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"042a070c38c11ce36b1152c6ffff9004025a1612d126acb29f4fe7cbdd1ae01b","family_id":"qf-proof-counterexample-universal","seed":20261239,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"even\",\"universal_statement\":\"每個整數都是偶數\",\"witness\":9}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is even.” and the stated witness n=9. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=9 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0019","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1ae79a3b3c5f86ba527a11691ce7215c4e83a32702e8cc2ac2a5c37e69b5ea8","family_id":"qf-proof-counterexample-universal","seed":20261240,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"positive\",\"universal_statement\":\"每個整數都大於 0\",\"witness\":-8}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than 0.” and the stated witness n=-8. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-8 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0020","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27f00ff492f12fa8874d0bbd22fcfe5f657d89f4ced907f3dcc19a7e2b1dd15f","family_id":"qf-proof-counterexample-universal","seed":20261241,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":100,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"greater\",\"universal_statement\":\"每個整數都大於 100\",\"witness\":100}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than 100.” and the stated witness n=100. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=100 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0021","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e1a4d2116e8b014f0f9f187a1270c03e1ef9f44387bf94d80918aa4c871c4ff","family_id":"qf-proof-counterexample-universal","seed":20261242,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":-2,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"less\",\"universal_statement\":\"每個整數都小於 -2\",\"witness\":-2}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is less than -2.” and the stated witness n=-2. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-2 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0022","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9b428e4110c34289df56aa09c74aee562efd7317c931cbc398baf9f4d933ad8","family_id":"qf-proof-counterexample-universal","seed":20261243,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":4,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 4 的倍數\",\"witness\":6}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 4.” and the stated witness n=6. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=6 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0023","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79d1e6520762f6a297ed27791cb005a70e1b45629a6f1914409a5425c469a241","family_id":"qf-proof-counterexample-universal","seed":20261244,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"zero\",\"universal_statement\":\"每個整數都等於 0\",\"witness\":12}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer equals 0.” and the stated witness n=12. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=12 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0024","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b55d97beaeb1eca704d458b83f3fae5721b0b6ff099ef99e13147110cce64b14","family_id":"qf-proof-counterexample-universal","seed":20261245,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":null,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"nonnegative\",\"universal_statement\":\"每個整數都大於或等於 0\",\"witness\":-12}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is greater than or equal to 0.” and the stated witness n=-12. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=-12 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20260921-proof-counterexample-universal-0025","batch_id":"batch-20260921-proof-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78c1a265989343c2e24a780b160203e7768a8a6cd5d3b0cdbd659de87c32769c","family_id":"qf-proof-counterexample-universal","seed":20261246,"semantic_template":"qf-proof-counterexample-universal/{\"bound\":9,\"counterexample_role\":\"falsifies_the_stated_universal_only\",\"domain\":\"integers\",\"predicate_id\":\"multiple\",\"universal_statement\":\"每個整數都是 9 的倍數\",\"witness\":10}"},"fields":{"title":"Counterexample to a Universal Proposition","stem":"The domain is explicitly the integers. Consider the universal proposition “Every integer is a multiple of 9.” and the stated witness n=10. Does this witness refute that universal proposition?","answer":{"kind":"choice","value":"Yes","choices":["Yes","No"],"explanation":"n=10 lies in the integer domain and fails the required predicate, so it is a valid counterexample to this stated universal proposition only."},"skills":["universal propositions","counterexample witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}