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NOTE: b' means $b$ not

I'm trying to convert ab'd + ab'cf to product of sums form

My professor gave us the following hint:

"Invert the equation, reduce it to sum-of-products, then invert it again. The result will be the original equation, but in product-of-sums form."

Now, I think I have the inverting portion pretty down pat. I just use De Morgan's laws.

So p' = (a'+ b + d') \cdot (a' + b + c' + f')

However, I'm not comfortable with the next step.

Does p' expand to: (a'a' + a'b + a'c' + a'f') + (ba' + bb + bc' + bf') + (d'a' + d'b + d'c' + d'f')\ ?

Ignoring the optimization for now...

Then p' =\ldots ?

Do I do this? p' = ( (a'a' + ba') + (a'a' + bb)\cdots and so on?

I feel like I'm doing this wrong.

Any help would be greatly appreciated. I'll respond quickly!

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Yes: multiplication distributes over products, so (a'+b+d')(a'+b+c'+f') = a'(a'+b+c'+f') + b(a'+b+c'+f') + d'(a'+b+c'+f'), and you can then distribute again each of the factors on the right.

You should then simplify in any number of ways; for example, you have a'b and ba', and since a'b+a'b = a'b, you can drop one of them. Since $bb=b$, you can rewrite $bb$ as $b$; etc.

In the end, you should have a sum of products. Then you can just invert.

For example, if at the end you had p' = a'b + bc' + d'f' + a'f' (you won't, but say you did), then you would have p = p'' = (a'b + bc' + d'f' + a'f')' = (a'b)'\cdot(bc')'\cdot (d'f')'\cdot (a'f')' and then you would apply De Morgan's laws to each of the factors, e.g., (a'b)' = a+b', so you would have p = (a+b') \cdot (b'+c)\cdot (d+f)\cdot (a+f), a product of sums.

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    Thanks again, you've made my day by removing that silly mental block :)2011-10-06