# Boolean algebra question: Converting between sum-of-products and product-of-sums

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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Thanks for making my post look pretty Arturo :) – user13327 Oct 6 '11 at 16:39

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.
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.
@Silver: Yes: in a boolean algebra, addition distributes over multiplication, and multiplication distributes over addition. However, what you are talking about is not "distributivity", it's associativity: $x+(y+z) = (x+y)+z$, and so we express the whole thing simply as $x+y+z$. So $$(a'+a'b)+(a'b+b) = a' + a'b + a'b + b = (a'+b)+(a'b+a'b) = (a'+a'b) + (a'b+a'b) = (a'+(a'b+(a'b+b))),$$ or associated any way you want. – Arturo Magidin Oct 6 '11 at 17:04