Computer Science Canada Abit of help simplifying this boolean statement? |
Author: | Macmee [ Sun Sep 23, 2012 10:54 am ] |
Post subject: | Abit of help simplifying this boolean statement? |
Hi! I'm attempting to simplify this boolean function by hand (for a discrete structures course): (P ∧ Q ∧ R) ∨ (~P ∧ Q ∧ R) ∨ (~P ∧ (~Q ∨ ~R) I know I can turn that into: (P ∧ Q ∧ R) ∨ (~P ∧ Q ∧ R) ∨ (~P ∧ ~Q) ∨ (~P ∧ ~R) and then combine (P ∧ Q ∧ R) and (~P ∧ ~Q) but the problem is I then get 6 different statements and if I keep combining things I just end up with a crazy amount of statements. I was wondering if there's a smarter way to go about simplifying large boolean statements as such then just combining everything and hoping you combined things in such a way that things cancel out. |
Author: | evildaddy911 [ Sun Sep 23, 2012 11:51 am ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
can you explain what you mean by ^, v and ~? |
Author: | Macmee [ Sun Sep 23, 2012 12:10 pm ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
My apologies: ~ = not / negation ^ = and / conjunction ∨ = or / disjunction |
Author: | Insectoid [ Sun Sep 23, 2012 12:18 pm ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
Well, first of all you've got (P^Q^R)v(~P^Q^R), which can be solved with common sense. The second part is a little trickier, and I don't have time to explain it right now. |
Author: | evildaddy911 [ Sun Sep 23, 2012 12:46 pm ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
~X^~Y = X ~v Y; you can apply this twice right now |
Author: | viperfan7 [ Mon Mar 18, 2013 1:51 pm ] | ||||
Post subject: | Re: Abit of help simplifying this boolean statement? | ||||
Necro post but no one posts in here as it is, so why not do something Well using truth tables I got it down to ~P v (Q ^ R) Table for (P ∧ Q ∧ R) ∨ (~P ∧ Q ∧ R) ∨ (~P ∧ (~Q ∨ ~R)
Table for ~P ∨ (Q ∧ R)
ALWAYS set up a truth table, they make life SO much easier and if you expand and simplify (P ∧ Q ∧ R) ∨ (~P ∧ Q ∧ R) ∨ (~P ∧ (~Q ∨ ~R)) -> (~P ∧ Q ∧ R) ∨ (P ∧ Q ∧ R) ∨ (~P ∧ (~Q ∨ ~R)) -> (~P ∧ Q ∧ R) ∨ (P ∧ Q ∧ R) ∨ (~P ∧ ~Q) ∨ (~P ∧ ~R) -> (~P ∧ Q ∧ R) ∨ (P ∧ Q ∧ R) ∨ ~(P ∧ Q) ∨ ~(P ∧ R) -> (~P ∧ Q ∧ R) ∨ (P ∧ Q ∧ R) ∨ ~(P ∧ Q ∧ R) -> ((P ∨ (~P ∧ Q ∧ R)) ∧ (Q ∨ (~P ∧ Q ∧ R)) ∧(R ∨ (~P ∧ Q ∧ R))) ∨ ~(P ∧ Q ∧ R) -> (1 ∧ (P ∨ Q) ∧ (P ∨ R) ∧ (Q ∨ ~P) ∧ Q ∧ (Q ∨ R) ∧ (R ∨ ~P) ∧ (R ∨ Q) ∧ R) ∨ ~(P ∧ Q ∧ R) -> (Q ∧ R) ∨ ~(P ∧ Q ∧ R) -> (Q ∧ R) ∨ (~P ∧ ~Q ∧ ~R) -> (~P ∨ (Q ∧ R)) ∧ (~Q ∨ (Q ∧ R)) ∧ (~R ∨ (Q ∧ R)) -> (~P ∨ Q) ∧ (~P ∨ R) ∧ (~Q ∨ Q) ∧ (~Q ∨ R) ∧ (~R ∨ Q) ∧ (~R ∨ R) -> (~P ∨ Q) ∧ (~P ∨ R) ∧ 1 ∧ (~Q ∨ R) ∧ (~R ∨ Q) ∧ 1 -> (~P ∨ Q) ∧ (~P ∨ R) ∧ (~Q ∨ R) ∧ (~R ∨ Q) -> (~P ∨ Q) ∧ (~P ∨ R) ∧ 1 -> (~P ∨ Q) ∧ (~P ∨ R) -> ~P ∨ (Q ∧ R) |
Author: | Insectoid [ Mon Mar 18, 2013 3:24 pm ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
That's so much work! There's only three parts to this problem, each of which simplifies easily with a little thought. (P ∧ Q ∧ R) ∨ (~P ∧ Q ∧ R) simplifies to Q^R, because, well, just look at it. So now we have Q^R ∨ (~P ∧ (~Q ∨ ~R) ~Q v ~R is the inverse of Q^R (there's some identity there) so we can just negate Q^R and get the same thing. Q^R v (~P^~(Q^R)) If Q^R evaluates to false, then ~(Q^R) is true. So we don't really need that bit. Which leaves us with (Q^R)v~P. |
Author: | viperfan7 [ Tue Mar 19, 2013 10:03 am ] |
Post subject: | RE:Abit of help simplifying this boolean statement? |
Well, its meant to show to process to get the answer, so that people can learn from it. That and I was bored. Making that post nicely formatted took longer then solving that thing. |