用真值表方法证明是否为重言式(p→q)∧(r→q)∧(p∨q)→q
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![用真值表方法证明是否为重言式(p→q)∧(r→q)∧(p∨q)→q](/uploads/image/z/5532654-30-4.jpg?t=%E7%94%A8%E7%9C%9F%E5%80%BC%E8%A1%A8%E6%96%B9%E6%B3%95%E8%AF%81%E6%98%8E%E6%98%AF%E5%90%A6%E4%B8%BA%E9%87%8D%E8%A8%80%E5%BC%8F%EF%BC%88p%E2%86%92q%29%E2%88%A7%28r%E2%86%92q%29%E2%88%A7%28p%E2%88%A8q%29%E2%86%92q)
用真值表方法证明是否为重言式(p→q)∧(r→q)∧(p∨q)→q
用真值表方法证明是否为重言式
(p→q)∧(r→q)∧(p∨q)→q
用真值表方法证明是否为重言式(p→q)∧(r→q)∧(p∨q)→q
p q p→q r q r->q (p→q)∧(r→q) p∨q (p→q)∧(r→q) ^(p∨q) 0 1 1 0 1 1 1 1 1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1 1 1 1 1 1 1 1 1 (p→q)∧(r→q) ^(p∨q) q (p→q)∧(r→q)∧(p∨q)→q 1 1 1 0 0 1 0 0 1 1 1 1 还有一种情况自己想着办 ..