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I'm a little stuck in my simplifying of this boolean logic expression. If it was $2 \times 2$, I know I could foil, but I can't find any law that will help me go any further. Would someone help me figure out where to go now?

In this attempt, $\cdot$ stands for logical AND, $+$ for logical OR, and $\overline{A}$ stands for NOT A.

Simplify: $\overline{(A+B)}\cdot\overline{(C+D+E)}+\overline{(A+B)}$ \begin{align} &\overline{(A+B)}\cdot\overline{(C+D+E)}+\overline{(A+B)}\\ \text{de Morgan's law}~~~&(\overline{A}\cdot\overline{B})\cdot(\overline{C}\cdot\overline{D}\cdot\overline{E})+\overline{(A+B)}\\ \text{de Morgan's law}~~~&(\overline{A}\cdot\overline{B})\cdot(\overline{C}\cdot\overline{D}\cdot\overline{E})+(\overline{A}\cdot \overline{B}) \end{align}

Here is an image of my attempt on paper.

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    $\begingroup$ Please use MathJax to write math correctly. Also, what operation is in between $\neg(A+B)$ and $\neg(C+D+E)$? $\endgroup$
    – manooooh
    Aug 31 '20 at 22:46
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    $\begingroup$ $U \wedge V \vee U = U$. $\endgroup$ Aug 31 '20 at 22:58
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    $\begingroup$ It's an and symbol. I'll look into MathJax now. Thank you. $\endgroup$ Sep 1 '20 at 1:48
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    $\begingroup$ When you refer to the expression been a 2 by 2, are you referring to it been solved on a Karnaugh map? $\endgroup$
    – user400188
    Sep 1 '20 at 2:07
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    $\begingroup$ I mean 2 x 2 by saying another way of writing something like (A + B)(C + D). $\endgroup$ Sep 1 '20 at 2:10
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You can just follow Eric Towers' comment:
Take $U = \overline{A+B}$ and $V = \overline{C+D+E}$, and apply one of the absorption laws: $$(U\cdot V)+U=U.$$ (The other would be $(U+V)\cdot U = U$.)
This was certainly what Michael Burr meant in his first comment.

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