Checking Boolean Algebra work - Simplification

Question

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 $

Answer 1

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.