• 2022-06-01
    把 "x ( P( x )®$ y Q( x ,y ))化为前束范式,推导过程正确吗? "x ( P( x )®$y Q( x ,y )) Û"x (¬ P( x ) ∨ $y Q( x ,y )) Û"x$y (¬ P( x ) ∨ Q( x ,y )) Û" x $y ( P( x ) ®Q ( x ,y ))
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    • 0

      【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)

    • 1

      ( )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) )

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      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))

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      表达式∀x(P(x,y)∨Q())∧∃y(R(x,y)→∀Zq())中∀x的辖域是() A: P(x,y) B: P(x,y)∨Q(z) C: R(x,y) D: P(x,y)∧R(x,y)

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      下列哪个不是谓词公式() A: ∀x﹁(P(x)→Q(x)) B: ∃y(∀x(P(x,y))→∀x(Q(x))) C: ﹁∃x(P(x)∨P(y)) D: ∃z∀x(∃y∨P(z))