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booleanboolean-logicboolean-expressionsimplifyboolean-operations

Simplifying Boolean Expression (A'BC) + (A'B'C) + (A'BC) + (AB'C)

please help me with simplifying this one. I am a bit new to these..

(A'BC') + (A'B'C) + (A'BC) + (AB'C)

the book i use shows and answer, which is,

Answer = A'B + B'C

I tried simplifying, but I get stucked with two eXors,
my simplification so far goes like this...

(A'BC') + (A'B'C) + (A'BC) + (AB'C)
A (BC' + B'C) + C (A'B + AB')

This doesn't seem to be a write way, Please someone help me simplify this,
and please show step by step, as I am sort of new..
Also I don't get how to simplify eXor further..

Solution

  • You have the Rule X' + X = True. SO

    (A'BC') + (A'B'C) + (A'BC)  + (AB'C) = 
(A'BC') + (A'BC)  + (A'B'C) + (AB'C) =   // just permuting the terms
A'B(C' + C)       + (A' + A)B'C      =   // factoring
A'B               + B'C