# Boolean-expression simplification $F = AB'C + (A'B' + ABC'D)'$

Here are my solutions. Hence: I am stuck on where or what path I am going to take

Problem: F = AB'C + (A'B' + ABC'D)'
Solution 1 ---------------
F = AB'C + (A'B' + ABC'D)'
= AB'C + (A'B')' (ABC'D)'
= AB'C + (A + B) (A' + B' + C + D')
= STUUUUCCKKK!!! -end

Solution 2 ---------------
F = AB'C + (A'B' + ABC'D)'
= AB'C + (A'B')' (ABC'D)'
= AB'C + (A + B) (A' + B' + C + D')
= AB'C + (C + D')
= AB'C + C + D'
= AB'C + D' - final answer


Which is which and I doubt all my answers. I don't know what to take as of the moment. Am I using all the theorems and laws correctly???

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I assume your non-present $\cdot$-multiplication is $\wedge$ (AND) and $+$ is $\vee$ (OR)? – AlexR Sep 23 '13 at 19:15

Moving from your third line and fourth line in solution 1, your fourth line is invalid: \begin{align} F & =AB'C + (A'B' + ABC'D)'\tag{1}\\ \\ & = AB'C + (A'B')' (ABC'D)'\tag{2}\\ \\ & = AB'C + (A + B) (A' + B' + C + D')\tag{3}\\ \\ & = AB'C + (AA'+AB'+AC+AD') (BA'+BB'+BC+BD')\tag{4}\end{align}

It should read \begin{align} F &= AB'C + (AA'+AB'+AC+AD') \color{red}{\bf +} (BA'+BB'+BC+BD')\tag{4}\\ \\ & = AB'C + AB' + AC + AD' + BA' + BC + BD'\tag{5}\\ \\ & = AB'(C + 1) + AC + AD' + BA' + BC + BD'\tag{6} \\ \\ & = AB' + AC + AD' + BA' + BC + BD' \tag{7}\\ \\ & = AB' + BA' + (A+B)(C + D')\tag{8}\end{align}

In solution $2$, you have \begin{align} F &= AB'C + (A'B' + ABC'D)' \tag{1}\\ \\ & = AB'C + (A'B')' (ABC'D)'\tag{2}\\ \\ & = AB'C + (A + B) (A' + B' + C + D') \tag{3}\\ \\ & = AB'C + (C + D')\tag{4} \\ \\ & = AB'C + C + D' \tag{5}\\ \\ & = AB'C + D' \tag{ final answer} \end{align} There's a problem with your move from line 3 to line 4, as well. Here, it seems that you aimed to distribute $(A+B)$, perhaps like so?: $$AB'C + (A + B)(A' + B') + (A + B)(C + D')\tag{4}$$ which is valid, but $(A + B)(A' + B') \neq (A + B)(A+B)'$, so you can not eliminate as a contradiction the term $(A + B)(A' + B')$. We can conclude from this expression that that $AB' + A'B$ holds. But then we have $$AB'C + (A + B)(A' + B') + (A + B)(C + D') \tag{4}$$ $$= AB'C + AB' + A'B + AC + AD' + BC + BD'\tag{5}$$ and now proceed as we did from step 5 in the first corrected proof.

One final note: Even if we had a statement from which we could obtain step $5$ in your second proof: $$F = AB'C + C + D'$$ then steps $(6)/(7)$ would be $$F = (AB' + 1)C + D' = 1C + D' = C + D'$$ You eliminated the wrong term in your final answer.

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Looks like you went to town and had fun! :-) +1 – Amzoti Sep 24 '13 at 1:31
So are you saying amWhy? that in this equation? There are two answers? I just want this to be simplified. I tried doing kmapping in electronics (this equation is electronics based). I mapped this and kmapped with it's results. The result is a 1. And I don't get it. – Sha Lala Sep 24 '13 at 10:44
If you will notice.. (A'B' + ABC'D)' .. I used demorgans theorem and came up with this. (A'B')' (ABC'D)' Which leads to this (A + B) (A' + B' + C + D').. – Sha Lala Sep 24 '13 at 10:52
Sha Lala - That's correct, it's the line after that (4) that was the problem. Your lines 1 - 3 are correct. Please read my post carefully. – amWhy Sep 24 '13 at 12:30