# Checking Boolean Algebra work - Simplification

I am currently working on an assignment for a CE class I am taking, and I wanted to know if I have been simplifying these equations correctly. I'm supposed to reduce them to a sum of products.

1) $(A+B)(C+B)(D’+B)(ACD’+E)$

$AC + AB +AD'+AB+AACD'+AE+CB+BB+D'B+BB+ACD'B + BE$ $AC + AB + AD' + ACD' + AE + CB + B + ACD'B + D'B + BE$
$AC + AB + AD' + ACD' + AE + CB + B + D'B + BE$
$AC + AB + AD' + AE + CB + B + D'B + BE$ $A(C + B + D' + E) + B(C + D' + E + B)$
$AC + AB + AD' + AE + B$ $AC + AD' + AE + B$

2). $(A’+B+C’)(A’+C’+D)(B’+D’)$

$A'A' + A'C' + A'D + A'B' + A'D' + A'B + BC' + BD + B'B + BD' + A'C' + C'C' + C'D + B'C'+ C'D'$
$A' + A'C' + A'D + A'B' + A'D' + A'B + BC' + BD + B'B + BD' + C' + C'D + B'C' + C'D'$
$A' + A'C + A' + A' + C' + B + B'B + C' + C'$ $A' + A' + A' + C' + B + B'B + C' + C'$
$A' + C' + B + B'B$ $A' + C' + B$

3). $[(AB’)+C’D]’$

$(AB')' (C'D)'$ $(A'+B)(C+D')$ $A'C + A'D' + CB + BD'$

4). $[A+B(C’+D)]’$

$A'B'+(C'+D)'$ $A'B' + CD'$

5). $(A \oplus BC)+BD+ACD$

$(A'BC + AB'C') + BD + ACD$
$A'BC + AB'C' + BD + ACD$ $A'B + A'C + AB'C' + BD + ACD$

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Improve the formatting like this – Sameh Shenawy Mar 15 '14 at 3:55
I fixed the formatting. I tried to fix XOR, but I don't know how to insert the XOR symbol. – user3000964 Mar 15 '14 at 4:13
use this \oplus for $\oplus$ and \veebar for $\veebar$ – Sameh Shenawy Mar 15 '14 at 4:27

My solution for 1) \begin{align} & (A+B)(C+B)(D'+B)(ACD'+E) \\ & (AC+B)(ACD'+D'E+BE) \\ & (ACD'+BE) \end{align} Rather than multiplying out everything and simplifying at the end, I have simplified intermediate factors to reduce the length of my calculation.

Simplification rules:
$$x + x = x$$
$$x + xy = x$$
$$x x = x$$
$$x x' = \text{false}$$
$$x + x' = \text{true}$$
$$x + x'y = x + y$$

It helps to keep factor literals in alphabetical order.

My solution for 2) \begin{align} & (A'+B+C')(A'+C'+D)(B'+D') \\ & (A'+BD+C')(B'+D') \\ & (A'B'+A'D'+B'C'+C'D') \end{align}

My solution for 3) \begin{align} & [(AB′)+C′D]′ \\ & (AB')'(C'D')' \\ & (A'+B)(C+D) \\ & (A'C+A'D+BC+BD) \end{align}

My solution for 4) \begin{align} & [A+B(C′+D)]′ \\ & (A'(B(C'+D))') \\ & (A'(B'+(C'+D)')) \\ & (A'(B'+CD')) \\ & (A'B'+A'CD') \end{align}

My solution for 5) \begin{align} & (A\oplus (BC))+BD+ACD \\ & (A'BC+A(BC)')+BD+ACD \\ & (A'BC+A(B'+C'))+BD+ACD \\ & (A'BC+AB'+AC')+BD+ACD \\ & (A'BC+AB'+AC'+BD) \end{align}

To verify the calculations, a truth table might make sense.

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