# 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)

The problem is, this doesn't seem to be right if I follow this tool: http://www.eecis.udel.edu/~roman/simplifier/Function.html

What exactly am I doing wrong here?

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Hint: Does $A'B+BC+ABC' = B(A'+C+AC') = B(A' + (C+C'A))$ lead to any simplification? – Dilip Sarwate Sep 6 '12 at 13:09
(C+C'A) = C+A, correct? I'm still getting the same result though. – Joao Ferreira Sep 6 '12 at 13:32
Perseverance, perseverance, perseverance! You got $B(A'+(C+C'A)) = B(A'+C+A)$, right? Is $A'+C+A$ the same as $A'+A+C$? Can you simplify that? – Dilip Sarwate Sep 6 '12 at 14:24
Ok, B(A′+A+C)= B(1+C) = B(1) = B; so now my Expression is AB' + B which results in A(B'+B) = A(1) = A... – Joao Ferreira Sep 6 '12 at 15:01
$AB' + B = A(B'+B)$? Which law are you using here? – Dilip Sarwate Sep 6 '12 at 15:26

$(A' + B')' + B(A' + AC) + ABC'$

$AB' + B(A' + C) + ABC'$

$AB' + BA' + BC + ABC'$

$A(B' + BC') + BA' + BC$

$A(B' + C') + BA' + BC$

$AB' + AC' + A'B + BC$

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.

$AB' + AC' + A'B + B$

$AB' + AC' + B$

$A + B + AC'$

$A + B$

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Cheers for your help, seems the online tool is giving out the wrong result then... – Joao Ferreira Sep 6 '12 at 16:45

$(A'+B)'+B(A'+AC)+ABC'=AB'+BA'+BAC+ABC'=AB'+BA'+AB(C+C')=AB'+BA'+AB=AB'+B(A+A')=AB'+B=A+B$

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(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 ]