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

问题

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..

回答1:

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

回答2:

I'll assume that multiplcation is AND, addition is OR and prime is negation.

Here's what I'd do:

1. (A'BC') + (A'B'C) + (A'BC) + (AB'C)
2. A'B(C'+C) + B'C(A'+A)
3. (C'+C) = 1 and (A'+A) = 1
4. A'B + B'C

Q.E.D.

回答3:

for simplifying boolean expressions use karnaugh maps. i think it is very much useful if we less number of variables. but if we have more variables then we can follow methods because this method is not that preferable.

(A'BC') + (A'B'C) + (A'BC) + (AB'C)    
answer just arrange the terms like this    
step 1:A'BC'+A'BC+AB'C+A'B'C    
now get common terms out    
step 2 : A'B(C'+C)+B'C(A+A')    
step 3 : A'B.1+B'C.1    
step 4 : A'B+B'C    

来源：https://stackoverflow.com/questions/21861329/simplifying-boolean-expression-abc-abc-abc-abc