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I have the Boolean expression: F = A'B'C'D + A'BC'D' + ABC + AB'C'D' + ABCD'. Note that the ' indicates the negation of the variable by my convention.

I am trying to show that F = BC + A'C' + B'D' is a logical solution from the above.

I first sat down and tried to think about a way to approach it, as I am some years removed from tackling them. It seemed out of the question to attempt a truth table for something so large, so I resolved to try a Karnaugh Map. However, I'm a little confused as to how to approach the ABC term that has no D. How do I include it in the Karnaugh Map? Is it a "don't care" term? Is there a simpler way I could be approaching things?

Any pointers in the right direction would be appreciated!

EDIT: There is additional information I was confused about. Beneath F(A,B,C,D) is:

D(A,B,C,D): Σ(0,2,5,6,7,10,11)

I thought this was a separate function, but now it seems like these are the don't care conditions for the above. I have followed the advice of expanding the three variable term and added the don't cares to my table and I think I have proven the solution.

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  • $\begingroup$ You can treat that $ABC$ term as $ABCD+ABCD'$. This is not the same as a "don't care" situation, when in that case $F$ can be $1$ or $0$ up to your choice. Back to our case, both $ABCD$ and $ABCD'$ terms have to be $1$. $\endgroup$
    – peterwhy
    Commented Jan 11, 2014 at 16:50
  • $\begingroup$ Why is truth table out of question? For example, A=D=False and B=C=True shows the identity is not true $\endgroup$
    – user114628
    Commented Jan 11, 2014 at 17:13

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I got the Karnaugh Map as this:

$$\begin{array}{r|c|c|c|c} CD\backslash AB&00&01&11&10\\\hline 00&0&1&0&1\\\hline 01&1&0&0&0\\\hline 11&0&0&1&0\\\hline 10&0&0&1&0 \end{array}$$

And this does not look like what you are supposed to show. Is the question correct?

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