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.

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)