简化布尔表达式(x + y)。(x + z)

时间:2017-07-21 04:46:18

标签: boolean boolean-expression

简化布尔表达式" (x + y)。(x + z)"

我认为答案是" x + y.z"但我不知道怎么做到。

4 个答案:

答案 0 :(得分:1)

你应该使用De Morgan Law(A + B)=(A' .B')。它以这种方式工作: (X+Y)=X'.Y' and (X+Z)=X'.Z' 通过交换:(X+Y).(X+Z)=(X'.Y').(X'.Z')=X'.Y'.X'.Z'=X'.X'.Y'.Z' 通过幂等性:X'.X'=X' 然后:X'.X'.Y'.Z'=X'.Y'.Z'=X'.(Y'.Z') 致电:Y'.Z'=W 然后:X'.(Y'.Z')=X'.W' 作者:De Morgan:X'.W'=(X+W) (I) 否定肯定:W'=Y'.Z' then W=(Y'.Z')'=Y'+Z'=Y.Z (II) 通过(I)和(II):(X+Y).(X+Z)=X+(Y.Z)=X+Y.Z

答案 1 :(得分:1)

(x+y)(x+z)-分发-> xx+xy+xz+yz -xx = x-> x+xy+xz+yz-> x+x(y+z)+yz -x = x.1-> x.1+x(y+z)+yz-> x(1+(y+z))+yz -1+(y + z)= 1-> x+yz

答案 2 :(得分:0)

这是使用幂等定律(xx = x)和吸收定律(x+xy = x)的解决方案。

(x+y)(x+z) = xx+xz+xy+yz = x+yz

答案 3 :(得分:0)

(x+y)(x+z) 
= xx + xz + yx + yz 
= x + xz + yx + yz (since xx = x eg 0.0 = 0 , 1.1 = 1)
= x(1 + z + y) + yz
= x(1 + y) +yz (since 1 + z = 1 e.g 1+0 = 1 or 1+1 = 1)
= x(1) + yz (since 1 +y =1 as explained above)
= x + yz