Recent content by pirtu

Thank you very much! Now I got it: F' = A' + (AB' + B') + (AC' + C') >>>>: (AB' + B') = B´ ja (AC' + C') = C´

Am I going right ridection if I change it like this: F´ = (A + B´) * (A + C´) + A´ + B´ + C´ F´ = F = A + (B´C´) + A´ + B´ + C´ I don't know how to make B´and C´ as a common factor?

and new try, why this does not open to me?!? :) F = (AB’C + A(BC)’ + (ABC)’)’ F` = AB`C + A (B` + C´) + (A´ + B´ + C´) F´ = AB´C + AB´ + AC´ + (A´ + B´ + C´) F´ = AB´ (C + 1) + AC´ + (A´ + B´ + C´) >>>>: (C + 1) = 1 F´ = AB´ + AC´ + (A´ + B´ + C´) F´ = A (B´ + C´) + (A´ + B´ + C´) Thank you...

I´m confused. Tried to change "sums", here is my result then: F = (AB’C + A(BC)’ + (ABC)’)’ F` = AB`C + A (B` + C´) + (A´ + B´ + C´) F´ = AB´C + AB´ + AC´ + (A´ + B´ + C´) F´ = AB´ (C + 1) + AC´ + (A´ + B´ + C´) F´ = AB´C + AC´ + (A´ + B´ + C´) F´ = A (B`C + C´) + (A´ + B´ + C´) F` = A (C´ B´) +...

tried something different but still no, can´t understand. F = (AB’C + A(BC)’ + (ABC)’)’ F´ = AB´C + AB´C´ + A´B´C´ F` = AB´ (C + C´) + A´B´C´ F´ = AB´ + A´B´C´ F´ = B´ (A + A` C`) F` = B` (A + C`) F = B (A´ + C)

Here is my "last try" so far: F = (AB’C + A(BC)’ + (ABC)’)’ F = A` B``C` + A`B``C`` + A``B``C`` F = A`BC` + A`BC + ABC F = A´BC` + BC (A´ + A) >>>>: A` + A = 1 F = BC + A`BC´ F = B (C + A´C´)

Can someone help how to simplyfy this boolean algebra? F = (AB’C + A(BC)’ + (ABC)’)’ answers should be ABC but I can`t make it, my calculations gives always wrong answer.