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I have a Boolean expression. we'll call it F.

for instance, F = ab' + ad + c'd + d'.

Assuming I did all the necessary steps too get F complement , i.e. F'.

I got: F' = b'd + ac'd'.

How do I get the Product of sums form of F?

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up vote 4 down vote accepted


(Note: I did not check your $F'$.)

Because of the way the De Morgan laws work, the complement of a product of sums is always a sum of products, and the complement of a sum of products is always a product of sums.

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Thank you, but that is exactly what I've got (on my paper...) and it doesn't work .. are you sure it's right? – Billie Feb 3 '13 at 8:25
@user1798362: I’m sure that it would be right if your $F'$ were right. The problem is that your $F'$ is wrong: I get $a'b'cd+ab'c'd'$ for $F'$. – Brian M. Scott Feb 3 '13 at 20:30

F = ab' + ad + c'd + d'

F'= (ab' + ad + c'd + d')' = (ab')' . (ad)' . (c'd)' . (d')' ---> a'.b' =(a'+b') De Morgans law = (a'+b)(a'+b')(c+d')(d)

Hence Product of sums.

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Hello, welcome to Math.SE. Thank you for your answer! It is, however, rather difficult to read. For some basic information about writing maths at this site see e.g. here, here, here and here. – Lord_Farin Oct 24 '13 at 7:00
I've been asking it a 1 year ago .. as you can guess it's not relevant anymore – Billie Oct 24 '13 at 10:24

If the boolean expression is AB+BC.Then the product of sum expression is (A+B)(B+C).To find this, the complement of sum product expression is the product of sum expression. (AB+BC)'=A'B'+B'C' By applying DEMORGANS LAW =(A'+B')(B'+C') Now, (A'+B')'(B'+C')'=(A+B)(B+C).

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