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so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?

$$ (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)(A+E) $$

Matti P.
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ali farhad
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  • I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached? – Matti P. Nov 20 '18 at 11:35
  • i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started.. – ali farhad Nov 20 '18 at 11:39
  • Just use the law of distributivity. Its not a mess. – Wuestenfux Nov 20 '18 at 11:43
  • did not work or maybe I can't work it out. I'm not too sure @Wuestenfux – ali farhad Nov 20 '18 at 13:22
  • @MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. https://photos.app.goo.gl/L2Epwh2U1S67qBL68 – ali farhad Nov 20 '18 at 19:41

1 Answers1

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$$ \begin{split} & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)(A+E) \\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)A \\ +& (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)E \\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})A\bar{B} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})AD \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})\bar{B}E \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})DE\\ % --------------- = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)A\bar{B} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)AD \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)\bar{B}E \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)DE\\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)AD \bar{E}\\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)\bar{B}\underbrace{E\bar{E}}_{=0} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)D\underbrace{E\bar{E}}_{=0}\\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)AD \bar{E}\\ \end{split} $$ Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $\bar{A}$ and $\bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have $$ \begin{split} & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)AD \bar{E}\\ =& \bar{B}(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & \bar{B}(\bar{A}+\bar{C}+D)AD \bar{E}\\ =& (\bar{C}+D)A\bar{B}\bar{E} \\ + & \bar{B}(\bar{C}+D)AD \bar{E}\\ =& A\bar{B}\left\{ (\bar{C}+D)\bar{E} + (\bar{C}+D)D \bar{E}\right\} \\ =& A\bar{B}\bar{E}(\bar{C}+D) \end{split} $$ You might want to recheck the details on this one.

Matti P.
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