[Math] Simplify ABC+AB’C+A’BC+A’B’C+A’B’C’

boolean-algebra

  1. ABC+AB'C+A'BC+A'B'C+A'B'C',
  2. AC + A'BC + A'B'C + A'B'C',
  3. C(A+A'B+A'B') + A'B'C',
  4. C(A + A'(B+B') + A'B'C',
  5. C ( A + A') + A'B'C',
  6. C + A'B'C',

but true answer is C+A'B.
Help me, what I missed?

Best Answer

I think that the true answer is $C+A'B'$ and not $C+A'B$ as you suggest.

You did well, but could go on with:

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

To get hold of the situation you could also make a Venn-diagram on:

$$(A\cap B\cap C)\cup(A\cap B^{\complement}\cap C)\cup(A^{\complement}\cap B\cap C)\cup(A^{\complement}\cap B^{\complement}\cap C)\cup(A^{\complement}\cap B^{\complement}\cap C^{\complement})=$$$$\left[[(A\cap B)\cup(A\cap B^{\complement})\cup(A^{\complement}\cap B)\cup(A^{\complement}\cap B^{\complement})]\cap C\right]\cup[A^{\complement}\cap B^{\complement}\cap C^{\complement}]=$$$$C\cup[A^{\complement}\cap B^{\complement}\cap C^{\complement}]=C\cup(A^{\complement}\cap B^{\complement})$$

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