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

Question

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')
  = AB'C + (AA'+AB'+AC+AD') (BA'+BB'+BC+BD')
  = AB'C + (AB'+AC+AD')(BA'+BC+BD')
  = 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???

Answer 1

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

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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.

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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.