【简答题】【消耗次数:1】
求公式“(pú (qùr)) ? (púqúr)”的主析取范式。
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相关题目
【简答题】 公式“(pú (qùr)) ? (púqúr)”的主析取范式为:[填空1]
【简答题】 求公式“(?p?q) ? (?qúp)”的主析取范式。
【简答题】 求公式“? (p?q)”的主析取范式。
【简答题】 公式“? (p?q)”的主析取范式为:[填空1]
【多选题】 公式“(?p?q) ? (?qúp)”的主析取范式中极小项下标有:
①  0.
②  1.
③  2.
④  3.
【多选题】 公式“? (p?q)”的主析取范式中极小项下标有:
①  0.
②  1.
③  2.
④  3.
【简答题】 公式“(pùq) ú (?púr)”的主合取范式为:[填空1]
【多选题】 公式“(pùq) ú (?púr)”的主合取范式中极大项下标没有:
①  0.
②  2.
③  4.
④  6.
【单选题】 公式(P→Q)→(P→R)的成假指派为( ) (按(P, Q, R)的顺序)。
①  (0, 1, 0)
②  (1, 0, 0)
③  (1, 1, 0)
④  (1, 1, 1)
【判断题】 设前提: p ? ?q, r ? q. 结论: r ? ?p. 则推理的形式结构为: (p ? ?q) ù (r ? q) ? (r ? ?p).
①  正确
②  错误
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