How do I simplify Boolean simplification ABC + AB' . ( A' C') + A'BC?

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The discussion focuses on simplifying the Boolean expression ABC + AB' . (A' C') + A'BC. Participants express confusion about the multiplication of AB' and (A' C'), leading to the realization that AA' equals zero, which simplifies the expression. The simplification process reveals that the second term cancels out, resulting in the expression reducing to ABC + A'BC. Further simplification leads to the conclusion that the expression can be factored to BC(A + A'), ultimately simplifying to BC. The conversation emphasizes the importance of recognizing terms that cancel out in Boolean algebra.
adnankhuram
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ABC + AB' . ( A' C') + A'BC

I am confused in AB' . ( A' C')

the multiplication would be AA'B'C' OR AA'C'+A'B'C'
 
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adnankhuram said:
ABC + AB' . ( A' C') + A'BC

I am confused in AB' . ( A' C')

the multiplication would be AA'B'C' OR AA'C'+A'B'C'

It is AA'B'C'.

ehild
 
Abc + ab' . ( a' c') + a'bc

abc + aa'b'c' + a'bc

abc + b'c' + a'bc ( aa' = 0)

bc(a + b'c' +a')

bc (a+a'+b'c') as a+a' = 1

bc( b'c')

bb'cc'

0

is this is right
 
adnankhuram said:
Abc + ab' . ( a' c') + a'bc

abc + aa'b'c' + a'bc

abc + b'c' + a'bc ( aa' = 0)

As aa'=0 the second term cancels. You have abc+a'bc. Continue from here.

ehild
 
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