Boolean simplification....kind of stuck on a certain form of expression

Discussion in 'Homework Help' started by RyanKim, Nov 17, 2011.

  1. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Hi guys I was wondering if anyone could take a look at 2 expressions I have that I can't seem to simplify. Im basically one step away from the answer at both junctures but I can't logically apply a law that I know how to use to get the answer.

    The first equation that I've pretty much solved has boiled down to this:

    1.) A!B! + B!C + A! + B! + C! (Where ! means not. E.g A! = A not)

    Final answer suppose to be A! + B! + C!

    and secondly

    2.) Q! + R! + S! + R!S + QR!R!

    Final Answer suppose to be Q! + R! + S!


    I cant seem to figure out what to do next. Im not aware of any law that can be applied here...though I do notice in both expressions B! and R! may have something to do with it respectively?

    Thanks!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    A!B! + B!C + A! + B! + C!

    how about ...

    A!(B! + 1) + B!(C + 1) + C!

    what's next?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I don't think your second question partial solution is correct - if that's the given solution. Can you post the original question?
     
  4. justtrying

    Active Member

    Mar 9, 2011
    329
    347
    for the second part try factoring out R!

    Q!+S!+R!(1+S+QR!) =

    is there any law you can apply now? think of 1+X = X

    p.s. even before grouping, you can simplify R!R! - recall XX = X
     
  5. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Ahh thanks T_N_K I think I got the first bit...(X + 1) = 1. For the second part perhaps I should have added brackets but that was the given solution by my prof.

    Q! + R! + S! + (R!S + QR!R!)

    So I guess R!R! = R! therefore (R!S + QR!) = R! ? Then Q! + R! + S! + (R!) = Q! + R! + S! Since ( X + X = X)?

    f = ( (QRS) (R + ( S + (QR!) )! )!
     
    Last edited: Nov 18, 2011
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    A computer generated minimization gives me the solution

    X = not Q or not R or not S or not RS;

    which doesn't match up with the solution you have.
     
  7. RyanKim

    Thread Starter Member

    Sep 18, 2011
    37
    1
    Really now? Hmm thats odd...Well if ud like to see my profs solutions for them Ill upload a PDF of it.
     
  8. justtrying

    Active Member

    Mar 9, 2011
    329
    347
    you can also do this:

    !(QRS(R+!(S+Q(!R))) = !(QRS(R+!S(!Q+R)) = !(QRS(R+!S!Q+!SR) = !(QRS)

    if you are wondering about how I got the final answer, check out basic multiplication rules ;)
     
    Last edited: Nov 18, 2011
  9. justtrying

    Active Member

    Mar 9, 2011
    329
    347
    not quite...

    (R!S + QR!) cannot equal R! but Q!+R!+S!+R!S+QR! is valid. Then what happens if you factor out R! (see earlier post)
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    My mistake. Finger trouble - I didn't enter the equation correctly. The application now spits out your given answer.
     
Loading...