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Can anyone verify this. If I can wrong can you point me in the correction direction:

$$AB'C'+A'B'C+A'BC'+AB'C = B'(AC'+A'C+AC)+A'BC' \rightarrow B'(AC'+C)+A'BC' \rightarrow B'(C+A)+A'BC'\rightarrow $B'C+B'A+A'BC'$$ (Is this correct or is it possible to do this $A'BC'=AB'C$, by NOTing it)

which then let me go further

$B'C+B'A+A'BC'\rightarrow B'(C+A+AC) \rightarrow B'(C+(A(1+C))) \rightarrow **B'(C+A)**$ (or is this correct?)

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  • $\begingroup$ To check whether it's correct, try plugging in some values for $A,B,C$ before and after the simplification. $\endgroup$ – Greg Martin Oct 2 '15 at 4:22
  • $\begingroup$ what about the notion that I am make NOTing? I that possible in Boolean? $\endgroup$ – learnmore Oct 2 '15 at 4:28
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An easy method to find and check the solution is the Karnaugh map:

             AB
       00  01  11  10
      +---+---+---+---+
   0  | 0 | 1 | 0 | 1 |
C     +---+---+---+---+
   1  | 1 | 0 | 0 | 1 |
      +---+---+---+---+

The map depicts the four minterms of your expression. Adjacent $1$ blocks in the map which only differ in one variable can be merged to reduce the number of terms.

Example:

$$A B' C' + A B' C = (A B')(C + C') = A B'$$

This leads to:

$$A B' + B' C + A' B C'$$

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