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。
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