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So far I've got:

$A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$

$= A'C'D'(B' + B) + C'D(A'B + AB')$

$= A'C'D'(1) + C'D(A \;\text{ XOR }\; B)$

$= C'[A'D' + D(A \;\text{ XOR }\; B)]$

Did I do this correctly? Is there a simpler solution?

Thanks

K

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Did you use the results from your previous question to help in this case? –  abiessu Oct 11 '13 at 12:01
    
The result in the previous question was great, but when mapped, used the same amount of gates (including nots) as the solution I included with my question. I ended up keeping my solution. –  Karim Oct 11 '13 at 12:11
    
Also, the solution in the previous question that someone provided used Karnaugh maps. We haven't gone over Karnaugh maps in class, so I'm not familiar with using them. –  Karim Oct 11 '13 at 12:12
    
Then I think this solution is similarly correct. :-) –  abiessu Oct 11 '13 at 12:16
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1 Answer

up vote 5 down vote accepted

Yes, your work is correct.

Simplification of a Boolean expression depends on context, and what form you are seeking in your "simplification": for example, conjunctive normal form (product of sums) or disjunctive normal form (sum of products), etc. Typically, one does not introduce "xor" $\oplus$ unless expressing the entire function in terms of $\oplus$, $\land$, $'$.

See, for example, the following, without the use of $\oplus$ (xor):

$$\begin{align} A'C'D'(B' + B) + A'BC'D + AB'C'D &= A'C'D' + A'BC'D + AB'C'D \\ \\ &= C'[A'D' + D(A'B + AB')]\end{align}$$

But again, all your manipulations are indeed correct.

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This needs a TU! +! –  Amzoti Oct 11 '13 at 12:56
    
You can't generally express Boolean functions in terms of xor and not alone -- perhaps you meant xor, not and and? –  Henning Makholm Oct 11 '13 at 13:30
    
@Henning Thank you. Yes, indeed, that's what I meant! I've edited accordingly. –  amWhy Oct 11 '13 at 13:33
    
Thank you for the help! You guys rock. –  Karim Oct 11 '13 at 14:18
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