Boolean algebra: I don't understand how the answer was achieved

1. Feb 11, 2012

keither

1. The problem statement, all variables and given/known data
Factor to obtain a product of sums. (Simplify where possible.)
BCD + C'D' + BC'D + CD

2. Relevant equations
Distributive: X(Y+Z)=XY+XZ and X+YZ=(X+Y)(X+Z)

Simplification: XY+XY'=X and (X+Y)(X+Y')=X
X+XY=X and X(X+Y)=X
(X+Y')Y=XY and XY'+Y=X+Y

Factoring/multiplying out: (X+Y)(X'+Z)=XZ+X'Y and XY+X'Z=(X+Z)(X'+Y)

Consensus: XY+YZ+X'Z=XY+X'Z and (X+Y)(Y+Z)(X'+Z)=(X+Y)(X'+Z)

3. The attempt at a solution
Here's what I've done so far:

BCD + C'D' + BC'D + CD Factor out D from BCD and CD

D(BC + C) + C'D' + B'C'D Factor out C from BC + C

D[C(B + 1)] + C'D' + B'C'D B + 1 = 1

DC + C'D' + B'D Factor out D from DC and B'C'D

D(C + B'C') + C'D' C + B'C' follows XY' + Y = X + Y

D(B' + C) + C'D' Distribute D

B'D + CD + C'D'

Attempt to simplify/factor further

B'D + CD + C'D' Add consensus of B'D and C'D' = B'C'

B'D + CD + C'D + B'C' Factor out D from B'D + CD, and C' from C'D + B'C'

D(B' + C) + C'(D + B') Use associate law to make C' + D

(C' + D)(B' + C)(D + B') Foil (B' + C)(D + B')

(C' + D)(B'D' + B' + CD' + B'C) Factor out B' from B'D' + B' + B'C

(C' + D)[B'(D' + 1 + C) + CD'] D' + 1 = 1

(C' + D)[B'(1 + C) + CD'] C' + 1 = 1

(C' + D)[B' + CD'] B' + CD' follows X + YZ = (X + Y)(X + Z)

(C' + D)(B' + C)(B' + D') Foil (B' + C)(B' + D')

(C' + D)(B' + B'D' + B'C + D'C) Redundant term B'D' gone due to consensus of B'C + D'C

(C' + D)(B' + B'C + D'C) Factor out C from (B'C + D'C)

(C' + D)[C(B' + D') + B'] I stopped here because I felt I was doing it wrong

The simplified product of sums is: (C' + D)(B' + C + D'). I don't understand how to get that answer from B'D + CD + C'D' . I think I'm using the laws and theorems incorrectly.