[Math] Simplify boolean expression $AB + A\bar B+ ABC $

boolean-algebra

Simplify $AB + A\bar B+ ABC $

I've been trying to simplify for a good while now. I'm using only the 10 rules but cannot find a way to simplify fully.

Best Answer

$AB + A\overline{B} + ABC = A(B + \overline{B}) + ABC = A + ABC = A(1 + BC) = A\overline{BC}$

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[Math] Boolean Simplification of A’B’C’+AB’C’+ABC’

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

It's that last step that used to trip me up. A'+AB = A'+B Forget what that law is called (identity?).

[Math] simplify boolean expression: xy + xy’z + x’yz’

After figuring things out with a Karnaugh map, we see that splitting up the $xy$ term will be helpful: \begin{align*} xy + xy'z + x'yz' &= xy(1) + xy'z + x'yz' \\ &= xy(z + z') + xy'z + x'yz' \\ &= (xyz + xyz') + xy'z + x'yz' \\ &= (xyz + xy'z) + (xyz' + x'yz') \\ &= xz(y + y') + yz'(x + x') \\ &= xz(1) + yz'(1) \\ &= xz + yz' \end{align*}

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[Math] Boolean Simplification of A’B’C’+AB’C’+ABC’

[Math] simplify boolean expression: xy + xy’z + x’yz’

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