[Math] Boolean function simplification

boolean-algebra

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?

Best Answer

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.

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

$AB' + AC' + B$

$A + B + AC'$

$A + B$

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