离散数学的谓词逻辑推理A1 = (∃x)(P(x)∧(∀y)(R(x,y)→L(x,y)))A2 = (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y)))B = ┐(∃x)(∀y)(R(y,x)∧Q(x))用逻辑推理法证明A1∧A2 => B对回答的老师万分感谢!前
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 07:07:49
![离散数学的谓词逻辑推理A1 = (∃x)(P(x)∧(∀y)(R(x,y)→L(x,y)))A2 = (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y)))B = ┐(∃x)(∀y)(R(y,x)∧Q(x))用逻辑推理法证明A1∧A2 => B对回答的老师万分感谢!前](/uploads/image/z/7063497-9-7.jpg?t=%E7%A6%BB%E6%95%A3%E6%95%B0%E5%AD%A6%E7%9A%84%E8%B0%93%E8%AF%8D%E9%80%BB%E8%BE%91%E6%8E%A8%E7%90%86A1+%3D+%28%26%238707%3Bx%29%28P%28x%29%E2%88%A7%28%26%238704%3By%29%28R%28x%2Cy%29%E2%86%92L%28x%2Cy%29%29%29A2+%3D+%28%26%238704%3Bx%29%28P%28x%29%E2%86%92%28%26%238704%3By%29%28Q%28y%29%E2%86%92%E2%94%90L%28x%2Cy%29%29%29B+%3D+%E2%94%90%28%26%238707%3Bx%29%28%26%238704%3By%29%28R%28y%2Cx%29%E2%88%A7Q%28x%29%29%E7%94%A8%E9%80%BB%E8%BE%91%E6%8E%A8%E7%90%86%E6%B3%95%E8%AF%81%E6%98%8EA1%E2%88%A7A2+%3D%3E+B%E5%AF%B9%E5%9B%9E%E7%AD%94%E7%9A%84%E8%80%81%E5%B8%88%E4%B8%87%E5%88%86%E6%84%9F%E8%B0%A2%21%E5%89%8D)
离散数学的谓词逻辑推理A1 = (∃x)(P(x)∧(∀y)(R(x,y)→L(x,y)))A2 = (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y)))B = ┐(∃x)(∀y)(R(y,x)∧Q(x))用逻辑推理法证明A1∧A2 => B对回答的老师万分感谢!前
离散数学的谓词逻辑推理
A1 = (∃x)(P(x)∧(∀y)(R(x,y)→L(x,y)))
A2 = (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y)))
B = ┐(∃x)(∀y)(R(y,x)∧Q(x))
用逻辑推理法证明A1∧A2 => B
对回答的老师万分感谢!
前面都看懂了,谢谢.但是想请教一下,最后一步有依据吗?我书上没看到说可以这么移否定词的.
离散数学的谓词逻辑推理A1 = (∃x)(P(x)∧(∀y)(R(x,y)→L(x,y)))A2 = (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y)))B = ┐(∃x)(∀y)(R(y,x)∧Q(x))用逻辑推理法证明A1∧A2 => B对回答的老师万分感谢!前
任何一本谈谓词逻辑的书均有量词转化法则!
(1)(∃x)(P(x)∧(∀y)(R(x,y)→L(x,y))) P(P规则)
(2) P(a)∧(∀y)(R(a,y)→L(a,y)) T(T规则) (1) ES(存在指定规则)
(3)P(a) T(2)
(4) (∀y)(R(a,y)→L(a,y)) T(2)
(5) (∀x)(P(x)→(∀y)(Q(y)→┐L(x,y))) P
(6) (P(a)→(∀y)(Q(y)→┐L(a,y))) T(5) US(全称指定规则)
(7) (∀y)(Q(y)→┐L(a,y))) T(3)(6)
(8) (R(a,b)→L(a,b)) T(4) US
(9) (Q(b)→┐L(a,b))) T(7) US
(10)L(a,b)→┐Q(b) T(9)
(11)R(a,b)→┐Q(b) T(8) (10)
(12)┐R(a,b)∨┐Q(b) T(11)
(13) (∃y)┐(R(y,b)∧Q(b)) T(12) EG(存在推广规则)
(14) (∀x)(∃y) ( ┐(R(y,x)∧Q(x))) T(13) UG(全称推广规则)
(15) ┐(∃x)(∀y)(R(y,x)∧Q(x)) T(14)