我试图找到函数的补充:X(Y+Z!W+!VS)
利用DeMorgan的定律。
我还需要将结果表示为产品总和。
我认为补充应该是:!X + (Y!(Z+!W)(V!S))
但我不确定......
然后,即使我,我也不确定如何将其纳入产品形式总和。
思想?
答案 0 :(得分:0)
![X(Y+Z!W+!VS)]
= ![XY + XZ!W + X!VS] distributive law
= !(XY)!(XZ!W)!(X!VS) De Morgan's law
= (!X+!Y)(!X+!Z+W)(!X+V+!S) De Morgan's law
= (!X!X+!X!Z+!X!W+!Y!X+!Y!Z+!YW)(!X+V+!S) distributive law
= (!X+!Y!Z+!YW)(!X+V+!S) x OR (x AND y) = x
= !X!X+!XV+!X!S+!Y!Z!X+!Y!ZV+!Y!Z!S+!YW!X+!YWV+!YW!S distributive law
= !X+!Y!ZV+!Y!Z!S+!YWV+!YW!S x OR (x AND y) = x
这是产品总和形式。更进一步......
= !X+!Y(!Z+W)(V+!S) distributive law
答案 1 :(得分:0)
要实施DeMorgans,通常最容易将事情分解成小部分
result=!(XA), (A= Y+Z!W+!VS) Simplify!
!X + !A DeMorgans
A = Y+B (B= Z!W+!VS) Simplify!
!A = !Y!B DeMorgans
B= C+D (C= Z!W D= !VS) Simplify!
!B = !C!D DeMorgans
!C = !Z+W DeMorgans
!D = V+!S DeMorgans
!B = (!Z+W)(V+!S) Put it back together!
!A = !Y(!Z+W)(V+!S)
result = !X+!Y(!Z+W)(V+!S)
要进入Sum of Products,需要更多代数。
!X already ok
+ !Y(!Z+W)(V+!S) not ok, let's call it !Y!B again.
!B= (!Z+W)(V+!S) remember this?
!B= !ZV+!Z!S+WV+W!S distribute!
-> !X + !Y(!ZV+!Z!S+WV+W!S) put it back together... still not quite right
!X + !Y!ZV + !Y!Z!S + !YWV + !YW!S distribute again!
现在我们有一个产品总和: !X +!Y!ZV +!Y!Z!S +!YWV +!YW!S