润哥人工智能习题-推导规约

1. 把下面的谓词公式化成子句集

(∀x)(∃y)(P(x,y)∨Q(x,y)→R(x,y))

解：

(∀x)(∃y)(P(x,y)∨Q(x,y)→R(x,y))

(∀x)(∃y)(P(x, y)∨(¬Q(x, y)∨R(x, y)))

(∀x)(P(x, f(x))∨¬Q(x, f(x))∨R(x, f(x)))

S={P(x, f(x))∨¬Q(x, f(x))∨R(x, f(x))} ​

2. 用归结反演法证明下面公式的永真性

[提示:先否定它，然后化成子句集，再实施归结和置换，得到最终的NIL(空子句)]

否定：

¬(∃x){[P(x)→P(A)]∧[P(x)→P(B)]}

¬(∃x){[¬P(x)∨P(A)]∧[¬P(x)∨P(B)]}

(∀x){[P(x)∧¬P(A)]∨[P(x)∧¬P(B)]}

(∀x){[P(x)∧¬P(A)]∨P(x)}∧{[P(x)∧¬P(A)]∨¬P(B)}

(∀x){P(x)∧[¬P(A)∨P(x)]∧[P(x)∨¬P(B)]∧[¬P(A)∨¬P(B)]}

(∀x){P(x)∧[¬P(A)∨P(x)]∧[P(x)∨¬P(B)]∧[¬P(A)∨¬P(B)]}

P(x)∧[¬P(A)∨P(x)]∧[P(x)∨¬P(B)]∧[¬P(A)∨¬P(B)]

子句集：

1. P(x1​)
2. ¬P(A)∨P(x2​)
3. P(x3​)∨¬P(B)
4. ¬P(A)∨¬P(B)