boolean - 布尔代数和德摩根定律

Question

我试图找到一个函数的补码：X(Y+Z!W+!VS)利用德摩根定律。

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，即使我不确定如何将其转换为产品总和形式。

想法？

Answer 1

  ![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

这是乘积之和的形式。再进一步...

= !X+!Y(!Z+W)(V+!S)                                    distributive law

Answer 2

要实现 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)

要将其纳入产品总和，需要更多的代数。

 !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