离散数学及其应用数理逻辑部分课后习题答案

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1、作业答案：数理逻辑部分P14：习题一1、下列句子中，哪些是命题？在是命题的句子中，哪些是简单命题？哪些是真命题？哪些命题的真值现在还不知道？（3）是无理数。答：简单命题，真命题。（9）吸烟请到吸烟室去！答：不是命题。（12）8是偶数的充分必要条件是8能被3整除。答：复合命题，假命题。14、讲下列命题符号化。（6）王强与刘威都学过法语。答：王强学过法语；刘威学过法语。符号化为：（10）除非天下大雨，他就乘班车上班。答：天下大雨；他乘班车上班。符号化为：（13）“2或4是素数，这是不对的”是不对的。答：2是素数；4是素数。符号

2、化为：15、设2+3=5.大熊猫产在中国。太阳从西方升起。求下列复合命题的真值。（2）（4）解答：真值为1；真值为1；真值为0.（2）真值为1；真值为1；真值为0；所以真值为0.（4）真值为1，真值为0，真值为1；所以真值为1.andapprovalpersonnelshallbestrictlyguidedbycreditpoliciesandrequirementsforreviewandapproval.Reviewapprovalthroughlogicalexamination,comparisonandveri

3、ficationofpersonnelaswellastheirownexperienceinreviewandapprovaltoenhancetheauthenticityofthecreditinformationaudit.Forobviouserrors,contradictions,orunreasonableorunexplainabledata,informationshouldbeindicatedinthereviewapproval.Contentreviewandapprovalpersonnels

4、houldconducta19、用真值表判断下列公式的类型。（4）0011111011011110010011100111所以为重言式。（7）0000111000110000101000011111010011101011000110100011111110000101001001101000110110101100111110110011101001111111所以为可满足式。andapprovalpersonnelshallbestrictlyguidedbycreditpoliciesandrequirementsf

5、orreviewandapproval.Reviewapprovalthroughlogicalexamination,comparisonandverificationofpersonnelaswellastheirownexperienceinreviewandapprovaltoenhancetheauthenticityofthecreditinformationaudit.Forobviouserrors,contradictions,orunreasonableorunexplainabledata,infor

6、mationshouldbeindicatedinthereviewapproval.ContentreviewandapprovalpersonnelshouldconductaP36:习题二3、用等值演算法判断下列公式的类型，对不是重言式的可满足式，再用真值表法求出其成真赋值。（1）解答：所以为永假式。（2）解答：所以因为永真式。（3）解答：为可满足式。真值表为000001001100010100011100100100101100110100111111andapprovalpersonnelshallbestric

7、tlyguidedbycreditpoliciesandrequirementsforreviewandapproval.Reviewapprovalthroughlogicalexamination,comparisonandverificationofpersonnelaswellastheirownexperienceinreviewandapprovaltoenhancetheauthenticityofthecreditinformationaudit.Forobviouserrors,contradiction

8、s,orunreasonableorunexplainabledata,informationshouldbeindicatedinthereviewapproval.Contentreviewandapprovalpersonnelshouldconducta4、用等值演算法证明下面的等值式。（2）解