Boolean algebra – prove $x_1 = x_2$

boolean-algebraproof-explanation

I'm trying to prove that two boolean algebra expressions are equivalent:

$x_1 = a'b'c + bc' + ac + ab'c$

$x_2 = b'c + bc’ + ab$

I got up to here:

LHS

$a'b'c + bc' + ac + ab'c$

RHS

$= (a + a')c + bc' + ab$

$= ab'c + a'b'c + bc' + ab$

$= a'b'c + bc' + ab'c + ab(c + c')$

$= a'b'c + bc' + ab'c + abc + abc'$

$= a'b'c + bc' + ab'c + (b + b')abc + abc'$

$= a'b'c + bc' + ab'c + abbc + ab'bc + abc'$

The next step is supposed to be

$= a'b'c + bc' + ab'c + ac + ab'c$

But I really can't see how they got to that step…

Someone please help me figure this one out and you will be a lifesaver thank you. Also let me know if I'm missing anything…

Best Answer

A Karnaugh map for your first expression is:

             bc
       00  01  11  10
      +---+---+---+---+
   0  | 0 | 1 | 0 | 1 |
a     +---+---+---+---+
   1  | 0 | 1 | 1 | 1 |
      +---+---+---+---+

You can convince yourself that the second expression has the same map.
Therefore, they are both equivalent.

A similar test would be to write a truth table for both expressions and compare the output columns.

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