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1、习题三求下列谓词公式的子句集。(1)xy(P(x,y)Q(x,y))解:去掉存在量词变为:P(a,b)Q(a,b)变成子句集{P(a,b),Q(a,b)}(2)xy(P(x,y)Q(x,y))解:去掉蕴涵符号变为:xy(¬P(x,y)Q(x,y))去掉全称量词变为:¬P(x,y)Q(x,y)变成子句集{¬P(x,y)Q(x,y)}(3)xy((P(x,y)Q(x,y))R(x,y))解:去掉蕴涵符号变为:xy(¬(P(x,y)Q(x,y))R(x,y))否定符号作用于单个谓词变为:xy((¬P(x,y)
2、¬Q(x,y))R(x,y))去掉存在量词变为:x((¬P(x,f(x))¬Q(x,f(x)))R(x,f(x)))去掉全称量词变为:(¬P(x,f(x))¬Q(x,f(x)))R(x,f(x)化合取范式为:(¬P(x,f(x))R(x,f(x))(¬Q(x,f(x))R(x,f(x))变元:(¬P(x,f(x))R(x,f(x)))(¬Q(y,f(y))R(y,f(y)))变成子句集{¬P(x,f(x))R(x,f(x)),¬Q(y,f(y))R(y,f(y))}(4)x(P(x)y(P(y)R(x,y))
3、)解:去掉蕴涵符号变为:x(¬(P(x)y(P(y)R(x,y)))去掉存在量词变为:x(¬(P(x)(P(f(x))R(x,f(x)))去掉全称量词变为:(¬(P(x)(P(f(x))R(x,f(x)))化合取范式为:(¬(P(x)P(f(x)))(¬(P(x)R(x,f(x)))变元:(¬(P(x)P(f(x)))(¬(P(y)R(y,f(y)))变为子句集:{¬(P(x)P(f(x)),¬(P(y)R(y,f(y))}(5)x(P(x)x(P(y)R(x,y)))解:去掉蕴涵符号变为:x(P(x)
4、x(¬P(y)R(x,y)))去掉存在量词变为:P(a)x(¬P(y)R(a,y))去掉全称量词变为:P(a)(¬P(y)R(a,y))变成子句集:{P(a),¬P(y)R(a,y)}(6)xyzuvw(p(x,y,z,u,v,w)(Q(x,y,z,u,v,w)¬R(x,z,w)))解:去掉存在量词变为:zv(p(a,b,z,f(z),v,g(z,v))(Q(a,b,z,f(z),v,g(z,v)¬R(a,z,g(z,v)))去掉全称量词变为:p(a,b,z,f(z),v,g(z,v))(Q(a,b,z
5、,f(z),v,g(z,v)¬R(a,z,g(z,v))变元:p(a,b,x,f(x),y,g(x,y))(Q(a,b,z,f(z),v,g(z,v)¬R(a,z,g(z,v))化成子句集:{p(a,b,x,f(x),y,g(x,y)),Q(a,b,z,f(z),v,g(z,v)¬R(a,z,g(z,v))}3.试判断下列子句集中哪些是不可满足的。(1)S={P(y)¬Q(y),¬P(f(x))Q(y)}解:(1)P(y)¬Q(y)(2)¬P(f(x))Q(z)(适当改名使子句之间不含相同变元利用归结原理:(3)P(y)¬P(f
6、(x))(1)(2){y/z}(4)T{f(x)/y}归结不出空子句,所以原子句集是可以满足的。(2)S={¬P(x)Q(x),¬Q(y)R(y),P(a),R(a)}解:(1)¬P(x)Q(x)(2)¬Q(y)R(y)(3)P(a)(4)R(a)利用归结原理判断(5)Q(a)(1)(3){a/x}(6)R(a)(2)(5){a/x}归结不出空子句,所以是可满足的子句集。(3)S={¬P(x)¬Q(y)¬L(x,y),P(a),¬R(z)L(a,z),R(b),Q(b)}解:(1)¬P(x)¬Q(y)¬L(x,y)(2)P(a)
7、(3)¬R(z)L(a,z)(4)R(b)(5)Q(b)利用归结原理来进行判断(6)¬Q(y)¬L(a,y)(1)(2){a/x}(7)L(a,b)(3)(4){b/z}(8)¬L(a,b)(6)(5){b/y}(9)Nil(8)(7)得到NIL所以原子句集不可满足。(4)S={P(x)Q(x)R(x),¬P(y)R(y),¬Q(a),¬R(b)}解:(1)P(x)Q(x)R(x)(2)¬P(y)R(y)(3)¬Q(a))(4)¬R(b)利用归结原理来判断(5)(6)(7)(5)S={P(x)Q(x),¬Q(y)R(y),¬P
8、(z)Q(z),¬R(u)}解:(1)P(x)Q(x)(2)¬Q(y)R(y)(3)¬P(z)Q(z)(4)¬R(u)利用归结原