Boolean function simplification I'm having problems with the following expression:
(A'+B)'+B(A'+AC)+ABC'
And here is what I tried to simplify:
AB' + B(A'+AC) + ABC' (De Morgan's)
AB' + B(A'+C) + ABC' (Identity)
AB' + A'B + BC + ABC' (Distribuitive)

Here I convert to Standard SOP Form (first term is missing C or C'; second term is missing C or C'; third term is missing A or A')
I run this on Karnaugh maps, and final result is: A+B
For the Identity rule, I read:
Boolean Simplification of A'B'C'+AB'C'+ABC'
(A'+AB = A'+B)
What exactly am I doing wrong here?
 A: We'll start with the most obvious simplifications.
$(A' + B')' + B(A' + AC) + ABC'$
$AB' + B(A' + C) + ABC'$
$AB' + BA' + BC + ABC'$
$A(B' + BC') + BA' + BC$
$A(B' + C') + BA' + BC$
$AB' + AC' + A'B + BC$
At this point the expression cannot be obviously reduced further. The secret is to use the consensus theorem to introduce redundancy. The consensus theorem states that $XY + X'Z + YZ = XY + X'Z$ If we set $XY = A'B$ and $X'Z = AC'$ then our expession becomes
$AB' + AC' + A'B + BC +BC'$
From here we can do the standard reductions.
$AB' + AC' + A'B + B$
$AB' + AC' + B$
$A + B + AC'$
$A + B$
A: $(A'+B)'+B(A'+AC)+ABC'=AB'+BA'+BAC+ABC'=AB'+BA'+AB(C+C')=AB'+BA'+AB=AB'+B(A+A')=AB'+B=A+B$
A: (A'+B)'+B(A'+AC)+ABC'
=  (A'+B)'+B(A'+C)+ABC'         by absorption [ A'+AC  =  A'+C ]
=  (A)(B')+B(A'+C)+ABC'         by DeMorgan [ (A'+B)'  =  (A)(B') ]
=  (A)(B')+(BA'+BC)+ABC'        by distributivity [ B(A'+C)  =  (BA'+BC) ]
=  AB'+BA'+BC+ABC'      by associativity
=  AB'+A'B+BC+ABC'      by commutativity
=  AB'+AC'+A'B+BC       by absorption [ AB'+ABC'  =  AB'+AC' ]
=  A'B+AB'+BC'+BC       by consensus [ AB'+AC'+A'B  =  A'B+AB'+BC' ]
=  A'B+AB'+B        by absorption [ BC'+BC  =  B ]
=  B+AB'        by absorption [ A'B+B  =  B ]
=  A+B      by absorption [ B+AB'  =  A+B ]

Answer:  A+B
Using tool at www.logicminimizer.com 
