2020年新编离散数学形成性考核作业7答案资料小抄(数理逻辑部分)名师精品资料

电大离散数学作业7

电大离散数学数理逻辑部分形成性考核书面作业

本课程形成性考核书面作业共3次，内容主要分别是集合论部分、图论部分、数理逻辑部分的综合练习，基本上是按照考试的题型（除单项选择题外）安排练习题目，目的是通过综合性书面作业，使同学自己检验学习成果，找出掌握的薄弱知识点，重点复习，争取尽快掌握。本次形考书面作业是第三次作业，大家要认真及时地完成数理逻辑部分的综合练习作业。

要求：将此作业用A4纸打印出来，手工书写答题，字迹工整，解答题要有解答过程，要求2010年12月19日前完成并上交任课教师（不收电子稿）。并在07任务界面下方点击“保存”和“交卷”按钮，以便教师评分。

一、填空题

1．命题公式()

P Q P

→∨的真值是 1 ．

2．设P：他生病了，Q：他出差了．R：我同意他不参加学习.则命题“如果他生病或出差了，我就同意他不参加学习”符号化的结果为P∨Q→R

．

3．含有三个命题变项P，Q，R的命题公式P Q的主析取范式是(P Q┐R)∨

Q R)

．

4．设P(x)：x是人，Q(x)：x去上课，则命题“有人去上课．”可符号化为

?x ( P ( x) ∧Q ( x))．

5．设个体域D＝{a, b},那么谓词公式)

(

)

xA?

∨

?消去量词后的等值式为(A(a)∨A(b))∨(B(a) ∧B(b))．

6．设个体域D＝{1, 2, 3}，A(x)为“x大于3”，则谓词公式(?x)A(x) 的真值为0 ．7．谓词命题公式(x)((A(x )B(x )) C(y))中的自由变元为y ．

8．谓词命题公式(?x)(P(x) →Q(x ) R(x，y))中的约束变元为x ．

三、公式翻译题

1．请将语句“今天是天晴”翻译成命题公式．

解：

设P：今天是天晴

则该语句符号化为P

2．请将语句“小王去旅游，小李也去旅游．”翻译成命题公式．

设P：小王去旅游，Q：小李也去旅游

则该语句符号化为P∧Q

3．请将语句“如果明天天下雪，那么我就去滑雪”翻译成命题公式．

解：设P：明天天下雪Q：我就去滑雪

则该语句符号化为P→Q

4．请将语句“他去旅游，仅当他有时间．”翻译成命题公式．

解：设P：他去旅游Q：他有时间

则该语句符号化为P→Q

5．请将语句“有人不去工作”翻译成谓词公式．

解：设P(x)：x是人Q(x)：x不去工作

则谓词公式为(?x)(P(x)∧Q(x))

6．请将语句“所有人都努力工作．”翻译成谓词公式．

解：设P(x)：x是人Q(x)：x努力工作

则谓词公式为(?x)(P(x)→Q(x))

四、判断说明题（判断下列各题，并说明理由．）

1．命题公式P P的真值是1．

不正确，┐P∧P的真值是0，它是一个永假式，命题公式中的否定律就是┐P∧P=F

2．命题公式P(P Q)P为永真式．

正确

可以化简┐P∧（P→┐Q）∨P=┐P∧（┐P∨┐Q）∨P=┐P∨P=1，所以它是永真式当然方法二是用真值表

3．谓词公式))

→

?是永真式．

→

xP?

)

(

)

正确

?x P(x)→(?y G(x,y)→?xP(x))

=?x P(x)→(┐?y G(x,y)∨?xP(x))

=?x P(x)→(?y(┐G(x,y))∨?xP(x))

=┐?x P(x)∨(?y(┐G(x,y))∨?xP(x))

=┐?x P(x)∨?y(┐G(x,y))∨?xP(x)

=┐?x P(x) ∨?xP(x)∨?y(┐G(x,y))

=1∨?y(┐G(x,y))

所以该式是永真式

4．下面的推理是否正确，请给予说明．

(1) (x)A(x)B(x) 前提引入

(2) A(y) B(y) US (1)

不正确，(1)中()x的辖域仅是A(x)，而不是A(x) B(x)

四．计算题

1．求P→Q∨R的析取范式，合取范式、主析取范式，主合取范式．

解：┐P∨(Q∨R）= ┐P∨Q∨R

所以合取范式和析取范式都是┐P∨Q∨R

所以主合取范式就是┐P∨Q∨R

所以主析取范式就是(P Q R) (P Q R) (P Q R) (P Q R) (P Q R) (P Q R) (P Q R)

2．求命题公式(P Q)(R Q) 的主析取范式、主合取范式．

解：(P Q)(R Q)= (P Q) (R Q)= (P Q) (R Q)

其中(P Q)= (P Q) (R R)= (P Q R) (P Q R)

其中(R Q)= (R Q) (P P)= (P Q R) (P Q R)

所以原式=(P Q R) (P Q R) (P Q R) (P Q R) =(P Q R) (P Q R) (P Q R)

= (P Q R) (P Q R) (P Q R)=m2m3m7这就是主析取范式

所以主合取范式为M0M1M4M5M6

可写为(P Q R)(P Q R) (P Q R) (P Q R) (P Q R)

3．设谓词公式()((,)()(,,))()(,)

?→?∧?．

x P x y z Q y x z y R y z

（1）试写出量词的辖域；

（2）指出该公式的自由变元和约束变元．

解：(1)量词?x的辖域为P(x,y) (z)Q(y,x,z)

量词z的辖域为Q(y,x,z)

量词y的辖域为R(y,x)

(2)P(x,y)中的x是约束变元，y是自由变元

Q(y,x,z)中的x和z是约束变元，y是自由变元

R(y,x)中的x是自由变元，y是约束变元

4．设个体域为D={a1, a2}，求谓词公式y xP(x,y)消去量词后的等值式；

)xP(x, a2)

解：y xP(x,y)= xP(x, a

=( P(a 1, a1)P(a2, a1))( P(a1, a2)P(a1, a2))

五、证明题

1．试证明(P(Q R))P Q与(P Q)等价．

证：(P→(Q∨?R))∧?P∧Q?(?P∨(Q∨?R))∧?P∧Q

?(?P∨Q∨?R)∧?P∧Q

?(?P∧?P∧Q)∨(Q∧?P∧Q)∨(?R∧?P∧Q)

?(?P∧Q)∨(?P∧Q)∨(?P∧Q∧?R)

??P∧Q（吸收律）

??(P∨?Q) （摩根律）

2．试证明(x)(P(x) R(x))(x)P(x) (x)R(x)．

证明：

(1) (x)(P(x) R(x))P

(2) P(a) R(a)ES(1)

(3) P(a)T(2)

(4) (x)P(x) EG(3)

(5) R(a)T(2)

(6) (x) R(x)EG(5)

(7) (x)(P(x) R(x))T(4)(6)

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