对谓词公式(∀x)((∃y)﹁P(x,y)∨(∃y)( Q(x,y) ∧﹁R(x,y)))化简可以得到包含哪几项的子句?
A: P(x,f(x))∨Q(x,g(x))
B: ﹁P(x,f(x))∨Q(x,g(x))
C: ﹁P(y,f(y))∨﹁R(y,g(y))
D: P(y,f(y))∨R(y,g(y))
A: P(x,f(x))∨Q(x,g(x))
B: ﹁P(x,f(x))∨Q(x,g(x))
C: ﹁P(y,f(y))∨﹁R(y,g(y))
D: P(y,f(y))∨R(y,g(y))
举一反三
- 3.4对下列各题分别证明G是否为F1,F2,…,Fn的逻辑结论:(1)F:(Ǝx)(Ǝy)(P(x,y)G:(ꓯy)(Ǝx)(P(x,y)(2)F:(ꓯx)(P(x)∧(Q(a)∨Q(b)))G:(Ǝx)(P(x)∧Q(x))(3)F:(Ǝx)(Ǝy)(P(f(x))∧(Q(f(y)))G:P(f(a))∧P(y)∧Q(y)(4)F1:(ꓯx)(P(x)→(ꓯy)(Q(y)→[img=1x1]17e0a6a55067d30.gif[/img]L(x.y)))F2:(Ǝx)(P(x)∧(ꓯy)(R(y)→L(x.y)))G:(ꓯx)(R(x)→[img=1x1]17e0a6a55067d30.gif[/img]Q(x))(5)F1:(ꓯx)(P(x)→(Q(x)∧R(x)))F2:(Ǝx)(P(x)∧S(x))G:(Ǝx)(S(x)∧R(x))
- ( )不是有效的推理。 A: 前提:("x)(~P(x)ÞQ(x)), ("x)~Q(x)结论:P(a) B: 前提:("x)(P(x)ÞQ) 结论:("x)P(x)ÞQ C: 前提:("x)(P(x)∨Q(x)), ("x)(Q(x)Þ~R(x)) 结论:($x)(R(x)ÞP(x)) D: 前提:("x)(P(x)Þ(Q(x)∧R(x))), ($x)(P(x)∧S(x))结论:("x)(R(x)∧S(x)) E: 前提:("x)($y)P(x, y)结论:("x)($y)($z)(P(x, y)∧P(y, z)) F: 前提:("x)P(x)∨("x)Q(x)结论:("x)(P(x)∨Q(x)) G: 前提:("x)(G(x)ÞH(x)),~($x)(F(x)∧H(x))结论:($x)F(x)Þ($x)G(x) H: 前提:("x)(H(x)ÞM(x))结论:("x)("y)(H(y)∧N(x, y)) Þ ($y)(M(y)∧N(a, y) )
- 谓词公式∀xP(x)→∀xQ(x)∨∃yR(y)的前束范式为 A: ∀x∀z∃y(P(x)→Q(z) ∨ R( y)) B: ∃x∀z∃y(P(x)→Q(z) ∨ R( y)) C: ∀x∃y(P(x)→Q(x) ∨ R( y)) D: ∃x∀y(P(x)→Q(x) ∨ R( y))
- 【单选题】公式(∀x)[P(x)∧Q(x, A) →(∃y)[R(x, y)∨S(y)]]中,∀x的辖域为 , ∃y的辖域为 。 A. P(x); R(x, y) B. P(x)∧Q(x, A); R(x, y) C. P(x)∧Q(x, A)→(∃y)[R(x, y)∨S(y)]; R(x, y) D. P(x)∧Q(x, A)→(∃y)[R(x, y)∨S(y)]; R(x, y)∨S(y)
- 与公式("x)(P(x)∧Q(x, y))Þ($x)R(x, y)等值的是( )。 A: ("x)(P(x)∧Q(x, z))Þ($x)R(x, y) B: ("y)(P(y)∧Q(y, y))Þ($x)R(x, y) C: ("z)(P(z)∧Q(x, y))Þ($x)R(x, y) D: ("u)(P(u)∧Q(u, z))Þ($x)R(x, z)