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?
$$ (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)(A+E) $$
$$ \begin{split} & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)(A+E) \\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)A \\ +& (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})(\bar{B}+D)E \\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})A\bar{B} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})AD \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})\bar{B}E \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D+\bar{E})DE\\ % --------------- = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)A\bar{B} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)AD \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)\bar{B}E \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)(C+D)DE\\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)AD \bar{E}\\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)\bar{B}\underbrace{E\bar{E}}_{=0} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)D\underbrace{E\bar{E}}_{=0}\\ = & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)A\bar{B}\bar{E} \\ + & (\bar{A} + \bar{B} + E)(\bar{A}+\bar{C}+D)AD \bar{E}\\ \end{split} $$ 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.