布尔代数习题答案.doc

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1、练习8.11．证明在布尔代数中a∨(a’∧b>=a∨b,a∧(a’∨b>=a∧b证明：a∨(a’∧b>=(a∨a’>∧(a∨b>分配律=1∧(a∨b>布尔代数的定义=a∨b布尔代数的定义第二个式子是第一个式子的对偶式，对第一个式子用对偶原理即可得到。2．证明：(1>(a∨b>∧(c∨d>=(a∧c>∨(b∧c>∨(a∧d>∨(b∧d>(2>(a∧b>∨(c∧d>=(a∨c>∧(b∨c>∧(a∨d>∧(b∨d>并推广到一般情况。证明：只需证明第一式，用对偶原理即得第二式。(a∨b>∧(c∨d>=((a∨b>∧c>∨((a∨b>∧d>分配律=((a∧c>

2、∨(b∧c>>∨((a∧d>∨(b∧d>>分配律=(a∧c>∨(b∧c>∨(a∧d>∨(b∧d>结合律推广到一般情况：(1>(a1∨a2∨…∨an>∧(b1∨b2∨…∨bn>=(a1∧b1>∨(a1∧b2>∨…∨(a1∧bn>∨(a2∧b1>∨(a2∧b2>∨…∨(a2∧bn>∨…∨(an∧b1>∨(an∧b2>∨…∨(an∧bn>∨b5E2RGbCAP(2>(a1∧a2∧…∧an>∨(b1∧b2∧…∧bn>=(a1∨b1>∧(a1∨b2>∧…∧(a1∨bn>∧(a2∨b1>∧(a2∨b2>∧…∧(a2∨bn>∧…∧(an∨b1>∧(an∨b2>∧…

3、∧(an∨bn>3.证明：(1>(a’∧c’>∨(b∧c>∨(a∧b’>=(a’∧b>∨(a∧c>∨(b’∧c’>证明：左式=(a’∧c’>∨(b∧c>∨(a∧b’>=(((a’∧c’>∨b>∧((a’∧c’>∨c>>∨(a∧b’>分配律=((a’∨b>∧(c’∨b>∧(a’∨c>∧(c’∨c>>∨(a∧b’>分配律=((a’∨b>∧(c’∨b>∧(a’∨c>>∨(a∧b’>分配律=((a’∨b>∧(c’∨b>∧(a’∨c>>∨(a∧b’>分配律=(((a’∨b>∧(c’∨b>∧(a’∨c>>∨a>∧(((a’∨b>∧(c’∨b>∧(a’∨c>>∨b

4、’>分配律p1EanqFDPw=((a’∨b∨a>∧(c’∨b∨a>∧(a’∨c∨a>>∧((a’∨b∨b’>∧(c’∨b∨b’>∧(a’∨c∨b’>>分配律DXDiTa9E3d=(c’∨b∨a>∧(a’∨c∨b’>布尔代数的定义右式=(a’∧b>∨(a∧c>∨(b’∧c’>=(((a’∧b>∨a>∧((a’∧b>∨c>>>∨(b’∧c’>分配律=(((a’∨a>∧(b∨a>>∧((a’∨c>∧(b∨c>>>∨(b’∧c’>分配律=((b∨a>∧(a’∨c>∧(b∨c>>∨(b’∧c’>分配律=(((b∨a>∧(a’∨c>∧(b∨c>>∨b’>∧((

5、(b∨a>∧(a’∨c>∧(b∨c>>∨c’>>分配律RTCrpUDGiT5/5=(((b∨a∨b’>∧(a’∨c∨b’>∧(b∨c∨b’>>>∧(((b∨a∨c’>∧(a’∨c∨c’>∧(b∨c∨c’>>>>分配律5PCzVD7HxA=(a’∨c∨b’>∧(b∨a∨c’>布尔代数的定义所以，左式=右式，即原式成立。(2>(a∧b>∨(b∧c>∨(c∧a>=(a∨b>∧(b∨c>∧(c∨a>证明：(a∧b>∨(b∧c>∨(c∧a>=(((a∧b>∨b>∧((a∧b>∨c>>∨(c∧a>分配律=(b∧((a∧b>∨c>>∨(c∧a>吸收律=(b∧((a

6、∨c>∧(b∨c>>>∨(c∧a>分配律=(b∧(a∨c>>∨(c∧a>交换律、吸收律=((b∧(a∨c>>∨c>∧((b∧(a∨c>>∨a>分配律=(((b∧a>∨(b∧c>>∨c>∧(((b∧a>∨(b∧c>>∨a>分配律=((b∧a>∨c>∧((b∧c>∨a>吸收律=((b∧a>∨c>∧((b∧c>∨a>吸收律=((b∨c>∧(a∨c>>∧((b∨a>∧(c∨a>>分配律=(a∨b>∧(b∨c>∧(c∨a>幂等律、交换律4.证明：如果a∧b=a∧c且a∨b=a∨c，则b=c.证明：b=b∨(a∧b>吸收律=b∨(a∧c>题目条件=(b∨a>∧(

7、b∨c>分配律=(a∨c>∧(b∨c>题目条件=(a∧b>∨c分配律=(a∧c>∨c题目条件=c吸收律练习8.21．构造命题布尔代数<{0,1},∨,∧,﹁,0,1>上的下列布尔函数的真值表(1>f(x,y,z>=x∧(y∨z>xyz(y∨z>x∧(y∨z>0000000110010100111010000101111101111111(2>f(x,y,z>=x∨(y∧z>xyzy∧zx∨(y∧z>000000010001000011111000110101110015/511111(3>f(w,x,y,z>=w∧y∧(x∨z>w,x,y,z0000

8、000100100011010001010110011110001001101010111100110111