1把以下合适公式化简为合取范式的子句集:(1)(x)(y)(z){P(x)(x)[Q(x,y)R(z)]}(2)(x)(y){{P(x)[Q(x)R(y)]}(y)[P(f(y))Q(g(x))]}(3)(x)(y){P(x)[Q(x)R(y)]}(y){[P(f(y))Q(g(y))](x)R(x)}(1)·Ø(x)(y)(z){P(x)(x)[Q(x,y)R(z)]}·Ø(x)(y)(z){ØP(x)(x)[ØQ(x,y)R(z)]}·(x)(y)(z){P(x)(x)[Q(x,y)ØR(z)]}·P(A)[Q(f(y,z),y)ØR(z)]·{P(A),Q(f(y,z),y),ØR(w)}(2)·(x)(y){{P(x)[Q(x)R(y)]}(y)[P(f(y))Q(g(x))]}·(x)(y){Ø{P(x)[Q(x)R(y)]}(y)[ØP(f(y))Q(g(x))]}·(x)(y){ØP(x)[ØQ(x)ØR(y)](w)[ØP(f(w))Q(g(x))]}·(x){ØP(x)[ØQ(x)ØR(h(x))](w)[ØP(f(w))Q(g(x))]}·[ØP(x)ØQ(x)ØP(f(w))Q(g(x))][ØP(x)ØR(h(x))ØP(f(w))Q(g(x))]·{ØP(x1)ØQ(x1)ØP(f(w1)Q(g(x1)),ØP(x2)ØR(h(w2))ØP(f(w2))Q(g(x2))}(3)·(x)(y){P(x)[Q(x)R(y)]}(y){[P(f(y))Q(g(y))](x)R(x)}·Ø(x)(y){P(x)[Q(x)R(y)]}(y){Ø[ØP(f(y))Q(g(y))](x)R(x)}·(x)(y){ØP(x)[ØQ(x)ØR(y)]}(w){Ø[ØP(f(w))Q(g(w))](v)R(v)}·{ØP(A)[ØQ(A)ØR(y)]}{[P(f(w))ØQ(g(w))]R(v)}·ØP(A){[ØQ(A)P(f(w))][ØQ(A)ØQ(g(w))][ØR(y)P(f(w))][ØR(y)ØQ(g(w))]}R(v)·{ØP(A)ØQ(A)P(f(w1))R(v1),ØP(A)ØQ(A)Q(g(w2))R(v2),ØP(A)ØR(y3)P(f(w3))R(v3),ØP(A)ØR(y4)Q(g(w4))Rv4)}2假设已知下列事实:1)小李(Li)喜欢容易的(Easy)课程(Course)。2)小李不喜欢难的(Difficult)课程。3)工程类(Eng)课程都是难的。4)物理类(Phy)课程都是容易的。5)小吴(Wu)喜欢所有小李不喜欢的课程。6)Phy200是物理类课程。7)Eng300是工程类课程。请用归结反演法回答下列问题:1)小李喜欢什么课程?2)证明小吴喜欢Eng300课程将已知事实形式化表示为合适公式:(1)(x)[Course(x)Easy(x)Like(Li,x)];(2)(x)[Course(x)ØEasy(x)ØLike(Li,x)];(3)(x)[Course(x)Eng(x)ØEasy(x)];(4)(x)[Course(x)Phg(x)Easy(x)];(5)(x)[Course(x)ØLike(x)Like(Wu,x)];(6)Course(Phy200)Phy(Phy200);(7)Course(Eng300)Eng(Eng300);·问题表示为以下合适公式(目标公式):(1)(x)[Coure(x)Like(Li,x)];(2)Like(Wu),Eng300);·将所有事实和对应于问题的目标公式取反加以化简,并标准化为合取范式子句集:(1)ØCourse(x1)ØEasy(x1)Like(Li,x1);(2)ØCourse(x2)Easy(x2)ØLike(Li,x2);(3)ØCourse(x3)ØEny(x)ØEasy(x3);(4)ØCourse(x4)ØPhy(x4)Easy(x4);(5)ØCourse(x5)Like(Li,x5)Like(Wu,x5);(6)Course(Phy200);(7)Phy(Phy200);(8)Course(Eng300);(9)Eng(Eng300);(10)目标公式(1)的取反:(1)ØCourse(x6)ØLike(Li,x6);(11)目标公式(2)的取反:(1)ØLike(Wu,Eng300);·解决问题(1)令(10)的取反为:Ask(x6)=Course(x6)Like(Li,x6)提取的问题回答为:Course(Phy200)Like(Li,Phy200)即小李喜欢Phy200课程.·解决问题(2)3.对于规则PQ,已知p(Q)=0.04,LS=100,LN=0.4,利用主观Bayes方法求出P(Q/P)和p(Q/ØP):O(θ/P)=LS*O(θ)=100*0.04/(1-0.04)=4.2P(θ/P)=O(θ/P))/(1+O(θ/P))=4.2/5.2=0.81O(θ/P)=LN¬*O(θ)=0.4*0.04/(1-0.04)=0.017P(θ/P)=O(θ/P)/(1+O(θ/P))=0.017/1.017=0.017¬¬¬4.在上题中,若P自身的确定性依赖P’,且有p(P)=0.05,规则P’P的LS=120,LN=0.3,用观Bayes方法求出P(θ/P')。(1).求P(P/P')O(P/P')=LS*O(P)=120*0.05/(1-0.05)=6.4P(P/P')=O(P/P')/(1+O(P/P'))=6.4/7.4=0.87(2).求P(θ/P')因为P(P/P')=0.87>p(P),根据P(θ/P')=0.04+(0.81-0.04)*(0.87-0.05)/(1-0.05)=0.04+0.66=0.705.在MYCIN中,设有如下规则:R1:IFE1THENH(0.8)R2:IFE2THENH(0.6)R3:IFE3THENH(-0.5)R4:IFE4AND(E5ORE6)THENE1(0.7)R5:IFE7ANDE8THENE3(0.9)在系统运行中已从用户处得CF(E2)=0.8,CF(E4)=0.5,CF(E5)=0.6,CF(E6)=0.7,CF(E7)=0.6,CF(E8)=0.9,求H的综合可信度CF(H)。解(1)求证据E4,E5,E6逻辑组合的可信度(2)根据规则R4,求CF(E1)(3)求证据E7,E8逻辑组合的可信度(4)根据规则R5,求CF(E3)(5)根据规则R1,求CF1(H)(6)根据规则R2,求CF2(H)(7)根据规则R3,求CF3(H)(8)组合由独立证据导出的假设H的可信度CF1(H),CF2(H)和CF3(H),得到H的综合可信度:6.设学生考试成绩的论域为{A,B,C,D,E},小王成绩得A、得B、得A或B的基本概率分别分配到0.2、0.1、0.3,Bel({C,D,E})为0.2;请给出Bel({A,B})、Pl({A,B})和f({A,B})。答:Bel({A,B})=m({A})+m({B})+m({A,B})=0.2+0.1+0.3=0.6Pl({A,B})=1-Bel({C,D,E})=1-0.2=0.8f({A,B})=Bel({A,B})+|{A,B}|/|U|·[Pl({A,B})-Bel({A,B})]=0.6+2/5·(0.8-0.6)=0.6+0.08=0.68