boolean logic: demorgan's theorem, NAND gates
<p>I am trying to build a circuit for the following portion of my boolean equation:</p>

<pre><code>(!A + !B + !C(!D + !E) 
</code></pre>

<p>Expanding this, we of course get: </p>

<pre><code>(!A + !B + !C!d + !C!E)
</code></pre>

<p>With demorgan's theorem, I thought we could reduce it to</p>

<pre><code>(ABCDCE)!
</code></pre>

<p>Which I though could be reduced to: 
ABCDE because A * A = A.</p>

<p>However, when I put in A B C D E into a NAND gate, I do not get the results that I need. Why is this? Where did I go wrong in my logic?</p>

<p>EDIT: so @Leandro Caniglia explained the problem:</p>

<pre><code>!A + !B + !C + !D + !E != !A + !B + !C!D + !C!E
</code></pre>

<p>So is the furthest I could reduce it:
<code>(AB)! + !C(DE)!</code> ? </p>

<p>Is there any way else that could simplify it?</p>