# Boolean Logic Design - Reduction

I have the following function to be reduced/simplified.

F(A,B,C,D) = BC + (A + C'D') where ' denotes the complement

```Here's my solution: = BC + (A + C'D')' = BC + (A + (C+D) = BC + (A + C + D) = BC + C + A + D = C(B + 1) + A + D = C*1 + A + D = C + A + D ```

Is this correct?

### Answer1:

As in traditional algebra, if you do something to one side of the equation, you must do it to the other side, including complementing. Here we state the original equation:

F'(A,B,C,D) = BC + (A + (CD)')

Since we have F' instead of F, my intuition tells me to complement both sides, but first I distribute the complement in the term (CD)' to make life easier in the long run:

F' = BC + (A + (C'+ D'))

Now we can complement both sides of the equation:

1: F = '(BC)'(A + (C'+ D')) The OR becomes AND after distributing complement

Now let's distribute the complements inside just to see what we get:

2: F = (B'+ C')(A'(CD))

Now we can just distribute the right term (A'(CD)) over the two terms being OR'ed:

3: F = B' (A'(CD)) + C' (A'(CD))

We see that the right term goes away since we have a CC' and thus we are left with:

4: F = A'B'CD

Hopefully I didn't make a mistake. I know you've found the answer, but others reading this might have a similar question and so I did it out to save duplicate questions from being asked. Good Luck!

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