# Reduce

• Oct 25th 2010, 05:44 AM
Zaph
Reduce
Hello, my head hurts. I'm trying to reduce this:

$\displaystyle M=A(B+C+B\bar{C})+ B(A \bar{C}+AB)+ BC+AB\bar{C}+(\bar{A}+C)$
the last part $\displaystyle (\bar{A}+C)$should also have a \bar around the whole thing but i couldnt get that to work in tex.

I got this far:

$\displaystyle AB + AC + BC + AB\bar{C} + AB\bar{C} B\bar{A}C + BAB + A + \bar{C} = B(A + C + A\bar{C} + A\bar{C} + \bar{A}C + AB) + AC + A + \bar{C}$

But im not sure if this is correct, if it is, hints for further reducing is very welcome :)
• Oct 25th 2010, 10:32 AM
emakarov
I'll write A' for $\displaystyle \bar{A}$, and similarly for other variables.

$\displaystyle \overline{\bar{A}+C}=A\bar{C}$. So, by using distributivity and eliminating the same minterms, or disjuncts (AB and twice ABC'), we get

AB + AC + ABC' + BC + AC'

Next, ABC' + AC' = AC'(B + 1) = AC', so we have

AB + AC + BC + AC'

AC + AC' = A, and A + AB = A, so the result is A + BC.
• Oct 25th 2010, 11:21 AM
Zaph
aha \overline did the trick :)

I've totally missed the obvious fact that X and X' eliminate eachother, it's been a long day, and i thank you :)