2

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?

Dragos Makovei
  • 25
Joao Ferreira
  • 165

3 Answers3

2

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.

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

$AB' + AC' + B$

$A + B + AC'$

$A + B$

Unreasonable Sin
  • 1,086
  • Cheers for your help, seems the online tool is giving out the wrong result then... – Joao Ferreira Sep 06 '12 at 16:45
  • Seems that there is an extra ' after the first B in the first line (i.e. should be (A' + B)', and not (A' + B')' as it is now). – obe Jun 04 '19 at 19:13
2

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

Aang
  • 14,672
0
(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

Mika
  • 151