屈婉玲版離散數(shù)學(xué)課后習(xí)題答案2
第四章部分課后習(xí)題參考答案3 .在一階邏輯中將下面將下面命題符號(hào)化,并分別討論個(gè)體域限制為(a),(b)條件時(shí)命題的真值:(1)對(duì)于任意x,均有三"-2=(x+%(x7T).(2)存在x,使得x+5=9.其中(a介體域?yàn)樽匀粩?shù)集合.(b)個(gè)體域?yàn)閷?shí)數(shù)集合.解:F(x):/-2=(x+')(x).G(x):x+5=9.(1)在兩個(gè)個(gè)體域中都解釋為xF(x),在(a)中為假命題,在(b)中為真命題。(2)在兩個(gè)個(gè)體域中都解釋為xG(x),在(a)(b)中均為真命題。4 .在一階邏輯中將下列命題符號(hào)化:(1)沒(méi)有不能表示成分?jǐn)?shù)的有理數(shù).(2)在北京賣菜的人不全是外地人.解:(1)F(x):x能表示成分?jǐn)?shù)H(x):x是有理數(shù)命題符號(hào)化為:x(F(x)H(x)(2)F(x):x是北京賣菜的人H(x):x是外地人命題符號(hào)化為:x(F(x)H(x)5 .在一階邏輯將下列命題符號(hào)化:(1)火車都比輪船快.(3)不存在比所有火車都快的汽車.解:(1)F(x):x是火車;G(x):x是輪船;H(x,y):x比y快命題符號(hào)化為:xy(F(x)G(y)H(x,y)(1)F(x):x是火車;G(x):x是汽車;H(x,y):x比y快命題符號(hào)化為:y(G(y)x(F(x)H(x,y)9 .給定解釋I如下:(a)個(gè)體域D為實(shí)數(shù)集合R.(b)D中特定元素孑=0.(c)特定函數(shù)f(x,y)=x-y,x,yD.(d)特定謂詞F(x,y):x=/(x,y):x<y,x,yD.說(shuō)明下列公式在I下的含義,并指出各公式的真值:(1) xy(G(x,y)F(x,y)(2) xy(F(f(x,y),a)G(x,y)答:(1)對(duì)于任意兩個(gè)實(shí)數(shù)x,y,如果x<y,那么xy.真值1.(2)對(duì)于任意兩個(gè)實(shí)數(shù)x,y,如果x-y=0,那么x<y.真值0.10 .給定解釋I如下:(a)個(gè)體域D=N(N為自然數(shù)集合).(b) D中特定元素有=2.(c) D上函數(shù)K也¥)=x+y,(x,y)=xy.(d) D上謂詞F(x,y):x=y.說(shuō)明下列各式在I下的含義,并討論其真值.(1) -xF(g(x,a),x)(2) -x-y(F(f(x,a),y戶F(f(y,a),x)答:(1)對(duì)于任意自然數(shù)x,都有2x=x,真值0.(2)對(duì)于任意兩個(gè)自然數(shù)x,y,使得如果x+2=y,那么y+2=x.真值0.11.判斷下列各式的類型:(1) 7-'-二(3)太芝;一yF(x,y).解:(1)因?yàn)閜(qp)p(qp)1為永真式;所以m0力-他區(qū)y)-F(xy)>為永真式;(3)取解釋I個(gè)體域?yàn)槿w實(shí)數(shù)F(x,y):x+y=5所以,前件為任意實(shí)數(shù)x存在實(shí)數(shù)y使x+y=5,前件真;后件為存在實(shí)數(shù)x對(duì)任意實(shí)數(shù)y都有x+y=5,后件假,此時(shí)為假命題再取解釋I個(gè)體域?yàn)樽匀粩?shù)N,F(x,y):x+y=5所以,前件為任意自然數(shù)x存在自然數(shù)y使x+y=5,前件假。此時(shí)為假命題。此公式為非永真式的可滿足式。13.給定下列各公式一個(gè)成真的解釋,一個(gè)成假的解釋。(1),(F(x)依就(2)三x(F(x)G(x)H(x)解:(1)個(gè)體域:本班同學(xué)F(x):x會(huì)吃飯,G(x):x會(huì)睡覺(jué).成真解釋F(x):x是泰安人,G(x):x是濟(jì)南人.(2)成假解釋(2)個(gè)體域:泰山學(xué)院的學(xué)生F(x):x出生在山東,G(x):x出生在北京,H(x):x出生在江蘇,成假解釋.F(x):x會(huì)吃飯,G(x):x會(huì)睡覺(jué),H(x):x會(huì)呼吸.成真解釋.第五章部分課后習(xí)題參考答案5.給定解釋I如下:(a)個(gè)體域D=3,4;(b)f(x)為f(3)4,f(4)3(c)F(x,y)為F(3,3)F(4,4)0,F(3,4)F(4,3)1.試求下列公式在I下的真值.(1)xyF(x,y)(3)xy(F(x,y)F(f(x),f(y)解:(1)xyF(x,y)x(F(x,3)F(x,4)(F(3,3)F(3,4)(F(4,3)F(4,4)(01)(10)1(2)xy(F(x,y)F(f(x),f(y)x(F(x,3)F(f(x),f(3)(F(x,4)F(f(x),f(4)x(F(x,3)F(f(x),4)(F(x,4)F(f(x),3)(F(3,3)F(f(3),4)(F(3,4)F(f(3),3)(F(4,3)F(f(4),4)(F(4,4)F(f(4),3)(0F(4,4)(F(3,4)F(4,3)(1F(3,4)(0F(3,3)(00)(11)(11)(00)12.求下列各式的前束范式。xF(x)yG(x,y)(5)XF(x1,x2)(H(xJx2G(x1,x2)(本題課本上有錯(cuò)誤)解:xF(x)yG(x,y)xF(x)yG(t,y)xy(F(x)G(t,y)xF(x1,x2)(H(x)x2G(x1,x2)xF(x1,x2)(H(x3)X2G(x3,x2)xF(x1,x4)x2(H(x3)G(x3,x2)x1*2任(不,人)(H(x3)G(x3,x2)15.在自然數(shù)推理系統(tǒng)F中,構(gòu)造下面推理的證明:(1)前提:xF(x)y(F(y)G(y)R(y),xF(x)結(jié)論:xR(x)(2)前提:x(F(x戶(G(a)AR(x),”F(x)結(jié)論:=x(F(x)AR(x)證明(1)xF(x)前提引入FEIxF(x)y(F(y)G(y)R(y)前提引入y(F(y)G(y)R(y)假言推理(FVG(c)戶R)UIFVG(c)附加R(c)假言推理xR(x)EG前提引入U(xiǎn)I假言推理化簡(jiǎn)合取引入xF(x)前提引入F(c)EI x(F(x戶(G(a)AR(x) F(c盧(G(a)AR(c) G(a)AR(c) R(c) F(c)AR(c) x(F(x)AR(x)