{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-counterexample"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-counterexample-01","title":{"zh-Hant":"反例與論域資格：n²+n+11 的質數性、失敗集合與最小反例","en":"Counterexamples and domain eligibility: primality of n²+n+11, the failure set and the least counterexample"},"summary":{"zh-Hant":"設 U={n∈ℤ ∣ 0≤n≤12}、V={n∈ℤ ∣ 0≤n≤9}、m(n)=n²+n+11、P(n)：「m(n) 是質數」。判定 φ_U：∀n∈U P(n)；求失敗集合 F_U 與最小反例；判定候選值 −11 的資格；並判定 φ_V：∀n∈V P(n)。","en":"Let U={n∈ℤ ∣ 0≤n≤12}, V={n∈ℤ ∣ 0≤n≤9}, m(n)=n²+n+11 and P(n): 'm(n) is prime'. Decide φ_U: ∀n∈U P(n); find the failure set F_U and the least counterexample; decide whether the candidate −11 is eligible; and decide φ_V: ∀n∈V P(n)."},"mko_ids":["mko-counterexample"]}],"title":"概念推導導讀"}