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电大离散考试模拟试题及答案
一、填空题1设集合AB,其中A={123}B={12}则A-B=____________________;A-B=__________________________.
2.设有限集合A|A|=n则|A×A|=__________________________.
3.设集合A={ab}B={12}则从A到B的所有映射是_______________________________________其中双射的是__________________________.
4.已知命题公式G=PQ∧R,则G的主析取范式是_________________________________________________________________________________________.
5.设G是完全二叉树,G有7个点,其中4个叶点,则G的总度数为__________,分枝点数为________________.6设A、B为两个集合A={124}B={34}则从AB=_________________________;AB=_________________________;A-B=_____________________.
7.设R是集合A上的等价关系,则R所具有的关系的三个特性是_____________________________________________________________________________.
8.设命题公式G=PQR,则使公式G为真的解释有__________________________,_______________________________________________________.
9.设集合A={1234}A上的关系R1={142332}R1={213243}则R1R2=________________________R2R1=____________________________R12=________________________.
10.设有限集AB,|A|=m|B|=n则||AB|=_____________________________.11设ABR是三个集合,其中R是实数集,A={x|-1≤x≤1xR}B={x|0≤x2xR}则A-B=__________________________B-A=__________________________A∩B=__________________________.
13.设集合A={23456},R是A上的整除,则R以集合形式列举法记为__________________________________________________________________.
14.设一阶逻辑公式G=xPxxQx,则G的前束范式是_______________________________.
15.设G是具有8个顶点的树,则G中增加_________条边才能把G变成完全图
16.设谓词的定义域为{ab}将表达式xRx→xSx中量词消除,写成与之对应的命题公式是__________________________________________________________________________.
17.设集合A={1234},A上的二元关系R={111223}S={132332}则RS=_____________________________________________________R2=______________________________________________________.
二、选择题1设集合A={2{a}34},B={{a}341},E为全集,则下列命题正确的是A{2}AB{a}AC{{a}}BED{{a}134}B.2设集合A={123}A上的关系R={1122233233},则R不具备.A自反性B传递性C对称性D反对称性3设半序集A≤关系≤的哈斯图如下所示,若A的子集B={2345}则元素6为B的A下界B上界C最小上界D以上答案都不对4下列语句中,是命题A请把门关上B地球外的星球上也有人Cx+56D下午有会吗?5设I是如下一个解释D={ab}则在解释I下取真值为1的公式是.AxyPxyBxyPxyCxPxxDxyPxy.
6.若供选择答案中的数值表示一个简单图中各个顶点的度,能画出图的是.A122345B123455C11123D
233456.
7.设G、H是一阶逻辑公式,P是一个谓词,G=xPxH=xPx则一阶逻辑公式GH是.A恒真的B恒假的C可满足的D前束范式.8设命题公式G=PQ,H=PQP,则G与H的关系是AGHBHGCG=HD以上都不是.9设AB为集合,当时A-B=B.AA=BBABCBADA=B=.10设集合A={1234}A上的关系R={11232434}则R具有A自反性B传递性C对称性D以上答案都不对11下列关于集合的表示中正确的为A{a}{abc}B{a}{abc}C{abc}D{ab}{abc}12命题xGx取真值1的充分必要条件是.A对任意x,Gx都取真值
1.B有一个x0,使Gx0取真值
1.C有某些x,使Gx0取真值
1.D以上答案都不对.
13.设G是连通平面图,有5个顶点,6个面,则G的边数是.A9条B5条C6条D11条.
14.设G是5个顶点的完全图,则从G中删去条边可以得到树.A6B5C10D
4.
15.设图G的相邻矩阵为,则G的顶点数与边数分别为.A45B56C410D
58.
三、计算证明题
1.设集合A={123468912},R为整除关系1画出半序集AR的哈斯图;2写出A的子集B={36912}的上界,下界,最小上界,最大下界;3写出A的最大元,最小元,极大元,极小元
1.设集合A={1234},A上的关系R={xy|xyA且xy}求1画出R的关系图;2写出R的关系矩阵.
2.设R是实数集合,是R上的三个映射,x=x+3x=2xx=x/4试求复合映射•,•••,••.
4.设I是如下一个解释D={23}abf2f3P22P23P32P3332320011试求1Pafa∧Pbfb;2xyPyx.
5.设集合A={1246812},R为A上整除关系1画出半序集AR的哈斯图;2写出A的最大元,最小元,极大元,极小元;3写出A的子集B={46812}的上界,下界,最小上界,最大下界.
6.设命题公式G=P→Q∨Q∧P→R求G的主析取范式
7.9分设一阶逻辑公式G=xPx∨yQy→xRx,把G化成前束范式.
9.设R是集合A={abcd}.R是A上的二元关系R={abbabccd}1求出rRsRtR;2画出rRsRtR的关系图.
11.通过求主析取范式判断下列命题公式是否等价1G=P∧Q∨P∧Q∧R2H=P∨Q∧R∧Q∨P∧R
13.设R和S是集合A={abcd}上的关系,其中R={aaacbccd}S={abbcbddd}.1试写出R和S的关系矩阵;2计算R•SR∪SR-1S-1•R-
1.
四、证明题
1.利用形式演绎法证明{P→QR→SP∨R}蕴涵Q∨S
2.设AB为任意集合,证明A-B-C=A-B∪C.
3.本题10分利用形式演绎法证明{A∨BC→BC→D}蕴涵A→D
4.本题10分AB为两个任意集合,求证A-A∩B=A∪B-B.参考答案
一、填空题
1.{3};{{3}{13}{23}{123}}.
2..
3.1={a1b1}2={a2b2}3={a1b2}4={a2b1};
34.
4.P∧Q∧R.
5.
123.
6.{4}{1234}{12}.
7.自反性;对称性;传递性.
8.
100101110.
9.{132231};{243342};{2233}.
10.2mn.
11.{x|-1≤x0xR};{x|1x2xR};{x|0≤x≤1xR}.
12.12;
6.
13.{2224263336445566}.
14.xPx∨Qx.
15.
21.
16.Ra∧Rb→Sa∨Sb.
17.{1322};{111213}.
二、选择题
1.C.
2.D.
3.B.
4.B.
5.D.
6.C.
7.C.
8.A.
9.D.
10.B.
11.B.
13.A.
14.A.
15.D
三、计算证明题
1.12B无上界,也无最小上界下界13;最大下界是
3.3A无最大元,最小元是1,极大元81290+;极小元是
1.
2.R={11212231323341424344}.
123.1•=x=x+3=2x+3=2x+
3.2•=x=x+3=x+3+3=x+63•=x=x+3=x/4+34•=x=x/4=2x/4=x/25••=••=•+3=2x/4+3=x/2+
3.
4.1Pafa∧Pbfb=P3f3∧P2f2=P32∧P23=1∧0=
0.2xyPyx=xP2x∨P3x=P22∨P32∧P23∨P33=0∨1∧0∨1=1∧1=
1.
5.12无最大元,最小元1,极大元812;极小元是
1.3B无上界,无最小上界下界12;最大下界
2.
6.G=P→Q∨Q∧P→R=P∨Q∨Q∧P∨R=P∧Q∨Q∧P∨R=P∧Q∨Q∧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∨P∧Q∧R∨P∧Q∧R=m3∨m4∨m5∨m6∨m7=
34567.
7.G=xPx∨yQy→xRx=xPx∨yQy∨xRx=xPx∧yQy∨xRx=xPx∧yQy∨zRz=xyzPx∧Qy∨Rz
9.1rR=R∪IA={abbabccdaabbccdd}sR=R∪R-1={abbabccbcddc}tR=R∪R2∪R3∪R4={aaabacadbabbbcbdcd};2关系图:
11.G=P∧Q∨P∧Q∧R=P∧Q∧R∨P∧Q∧R∨P∧Q∧R=m6∨m7∨m3=367H=P∨Q∧R∧Q∨P∧R=P∧Q∨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=m6∨m3∨m7=367GH的主析取范式相同,所以G=H.
13.12R•S={abcd}R∪S={aaabacbcbdcddd}R-1={aacacbdc}S-1•R-1={badc}.四证明题
1.证明{P→QR→SP∨R}蕴涵Q∨S1P∨RP2R→PQ13P→QP4R→QQ235Q→RQ46R→SP7Q→SQ568Q∨SQ
72.证明A-B-C=A∩~B∩~C=A∩~B∩~C=A∩~B∪C=A-B∪C
3.证明{A∨BC→BC→D}蕴涵A→D1AD附加2A∨BP3BQ124C→BP5B→CQ46CQ357C→DP8DQ679A→DD18所以{A∨BC→BC→D}蕴涵A→D.
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