Simplify Boolean Product of Sums Function

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I've got a product of sums expression: F=(A'+B+C')&(A+D')(C+D')

I need to show it as a sum of products and then simplify it. Right now I got: F=(A'&D')+(A&B&C)+(B&D')+(C&D')

But the problem is that the values in the truth table are not the same. I believe that I've made some mistakes when trying to show it as a sum of products and simplyfing.

Could you please help me out?

Thanks!

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1 Answer

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Your original expression is a product of sums:

(A' + B + C') (A + D') (C + D')

If you apply the and for the first two sums, you get:

(A'A + A'D' + AB + BD') (C + D')

A'A cancels out to false. In conjunction with the third sum, we get:

A'CD' + A'D'D' + ABC + ABD' + BCD' + BD'D'

Applying D'D' = D' gives us:

A'CD' + A'D' + ABC + ABD' + BCD' + BD'

A'CD' is covered by A'D' and can thus be omitted.

The minimized sum of products (the original six terms are covered by just four terms):

A'D' + C'D' + BD' + ABC

The terms of the expression shown in a Karnaugh-Veitch map:

enter image description here

The diagram helps to visually grasp which term is covered by which larger term.

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