Question

离散数学 归结推理法证明问题

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))
用归结推理法证明A1∧A2 => B
谢谢 请一定要用归结推理法

Answer 1

证明:
(1) ∃x(P(x)∧ ∀y(R(x,y)→L(x,y))) P
(2) P(a)∧ ∀y( R(a,y)→L(a,y) ) ES(1)
(3) ∀y( R(a,y)→L(a,y) ) T(2)I
(4) ∀x(P(x)→∀y(Q(y)→┐L(x,y)) ) P
(5) P(a)→∀y( Q(y)→┐L(a,y) ) US(4)
(6) P(a) T(2)I
(7) ∀y( Q(y)→┐L(a,y) ) T(5)(6)I
(8) R(a,b)→L(a,b) US(3)
(9) Q(b)→┐L(a,b) US(7)
(10) L(a,b)→┐Q(b) T(9)E
(11) R(a,b)→┐Q(b) T(8)(10)I
(12) ┐(R(a,b)∧Q(b) ) T(11)E
(13) ∃y┐(R(y,b)∧Q(b) ) EG(12)
(14) ∀x∃y┐(R(y,x)∧Q(x) ) UG(13)
(15) ┐∃yx∀y(R(y,x)∧Q(x) ) T(14)E