(1)(?x)(?A(x)?B(x)),(?x)?B(x)?(?x)A(x) 证明: (1)
(2) (3) (4) (5) (6)
(?x)(?A(x)?B(x))
P US,(1) P US,(3) T,(2),(4), EG,(5)
?A(a)?B(a) (?x)?B(x) ?B(a) A(a)
(?x)A(x)
(2)(?x)A(x)?(?x)B(x)?(?x)(A(x)?B(x))
证明: (1) ?(?x)(A(x)?B(x)) P(假设前提) (2) (?x)(?(A(x)?B(x))) T (3) (?x)(A(x)??B(x)) T (4) (?x)A(x)?(?x)?B(x) T (5) (?x)A(x) T (6) (?x)?B(x) T (7) (?x)A(x)?(?x)B(x) P
(8) (?x)B(x) T(5)(7) (9) ?B(a) ES(6) (10) B(a) US(8) (11) ?B(a)?B(a) T(9)(10) (12) (?x)(A(x)?B(x)) F(1)(11)
(3)(?x)(A(x)?B(x)),(?x)(C(x)??B(x))?(?x)(C(x)??A(x)) 证明: (1)
?(?x)(C(x)??A(x))
P(假设前提)
(2) (3) (4) (5) (6) (7) (8) (9)
(?x)?(?C(x)??A(x))
C(a)?A(a)
T,(1) US,(2) T,(3) T,(3) P US,(6) T,(5),(7) P US,(8) T,(4),(10) T,(8),(11)
C(a) A(a)
(?x)(A(x)?B(x)) A(a)?B(a) B(a)
(?x)(C(x)??B(x))
(10) C(a)??B(a) (11) ?B(a)
(12) B(a)??B(a)
(4)(?x)(A(x)?B(x)),(?x)(B(x)??C(x)),(?x)C(x)?(?x)A(x) 证明: (1) (?x)C(x) P (2) C(x) US(1) (3) (?x)(B(x)??C(x)) P (4) B(x)??C(x) US(3) (5) ?B(x) T(2)(4) (6) (?x)(A(x)?B(x)) P (7) A(x)?B(x) US(6) (8) A(x) T(5)(7) (9) (?x)A(x) UG(8) 2.用CP规则证明下列各式。
(1)(?x)(P(x)?Q(x))?(?x)P(x)?(?x)Q(x)
证明: (1) (?x)P(x) P(假设前提) (2) P(x) US(1) (3) (?x)(P(x)?Q(x)) P (4) P(x)?Q(x) US(3) (5) Q(x) T(2)(4) (6) (?x)Q(x) UG(5) (7) (?x)P(x)?(?x)Q(x) CP(1)(6) (2)(?x)(P(x)?Q(x))?(?x)P(x)?(?x)Q(x)
证明:由于(?x)P(x)?(?x)Q(x)??(?x)(?Q(x))?(?x)P(x)
?(?x)(?Q(x))?(?x)P(x)
因此,原题等价于证明(?x)(P(x)?Q(x))?(?x)(?Q(x))?(?x)P(x) (1) (?x)(?Q(x)) P(假设前提) (2) ?Q(x) US(1) (3) (?x)(P(x)?Q(x)) P (4) P(x)?Q(x) US(3) (5) P(x) T(2)(4) (6) (?x)P(x) UG(5) (7) (?x)(?Q(x))?(?x)P(x) CP(1)(6)
3.将下列命题符号化并推证其结论。
(1)所有的有理数是实数,某些有理数是整数,因此某些实数是整数。 解:首先定义如下谓词:
P(x):x是有理数 R(x):x是实数 I(x):x是整数
于是问题符号化为:
(?x)(P(x)?R(x)),(?x)(P(x)?I(x))?(?x)(R(x)?I(x))
推理如下:
(1) (?x)(P(x)?I(x)) P (2) P(a)?I(a) ES(1)
(3) (?x)(P(x)?R(x)) P (4) P(a)?R(a) US(3)
(5) P(a) T(2) (6) I(a) T(2) (7) R(a) T(4)(5) (8) R(a)?I(a) T(6)(7)
(9) (?x)(R(x)?I(x)) EG(8)
(2)任何人如果他喜欢步行,他就不喜欢乘汽车,每一个人或者喜欢乘汽车或者喜欢骑自行车,有的人不爱骑自行车,因而有的人不爱步行。 解:首先定义如下谓词:
P(x):x是人 F(x):x喜欢步行 C(x):x喜欢乘汽车 B(x):x喜欢骑自行车
于是问题符号化为:
(?x)(P(x)?F(x)??C(x)),(?x)(P(x)?C(x)?B(x)),
(?x)(P(x)??B(x))?(?x)(P(x)??F(x))推理如下:
(1) (?x)(P(x)??B(x)) P (2) P(a)??B(a) ES(1) (3) P(a) T(2)
(4) ?B(a) T(2) (5) (?x)(P(x)?C(x)? (6) P(a)?C(a)? (7) C(a)?B(x)) P
B(a) US(5)
B(a) T(3)(6)
(8) C(a) T(4)(7) (9) (?x)(P(x)?F(x)??C(x)) P
(10) P(a)?F(a)??C(a) US(9) F(a)) T(8)(10)
(11) ?(P(a)?(12) ?P(a)??F(a) T(11) (13) ?F(a) T(3)(12) (14) P(a)??F(a) T(3)(13) (15) (?x)(P(x)??F(x)) EG(14)
(3)每个科学工作者都是刻苦钻研的,每个刻苦钻研而且聪明的科学工作者在他的事业中
都将获得成功。华为是科学工作者并且他是聪明的,所以,华为在他的事业中将获得成功。 解:首先定义如下谓词:
P(x):x是科学工作者 Q(x):x是刻苦钻研的 R(x):x是聪明的
S(x):x在他的事业中将获得成功
定义个体a:华为 于是命题符号化为:
(?x)(P(x)?Q(x)),(?x)(P(x)?Q(x)?R(x)?S(x)),
P(a)?R(a)?S(a)推理如下:
(1) (?x)(P(x)?Q(x)) P (2) P(a)?Q(a) US(1)
离散数学答案陈志奎
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