1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Boolean algebra

  1. Apr 1, 2008 #1
    show that x'yc+xy'c+xy = yc + xc + xy




    3. The attempt at a solution

    - my first attempt at the solution was this:

    c(x'y+xy')+xc >>> c(y+x)+xc >>> yc+xc+xy

    I can't remember if that is valid or not. I seem to remember that A'B=B, but not sure. Can somebody please help me?
     
  2. jcsd
  3. Apr 1, 2008 #2
    Can you explain your notation? Specifically, what is c and ' ? I assume AND is denoted by nothing, and I figure ' is complement but I'm not entirely sure.
     
  4. Apr 1, 2008 #3
    - the ' is for a complement. x, y and c are just input variables (1 or 0). The "+" is for OR. so basically the problem ask to show how to reduce the left hand side to get whats on the right hand side using boolean algebra properties.
     
  5. Apr 1, 2008 #4
    Ok, well A'B is not the same as B. x'y + xy' is actually the definition of XOR, not OR, so you'll have to try something different. Try to use clever methods of ANDing always true statements or ORing ones that do not change the result (examples, xyc + xyc = xyc, xy=xy(c+c').
     
    Last edited: Apr 1, 2008
  6. Apr 1, 2008 #5
    alright. I remember now about using xy(c+c'), but I get stuck here...

    x'yc+xy'c+xyc+xyc' >>> yc(x+x')+xy'c+xyc' >>> yc + xy'c + xyc'

    I'm not real sure where to go from there.
     
  7. Apr 1, 2008 #6
    You haven't used the other property I identified. X+X=X. Don't simplify until you have all the terms of the form xy(c+c') (where x, y and c are interchanged somehow)
     
  8. Apr 1, 2008 #7
    I don't know...I'm lost here. Can you explain this in a little more detail:

    "Try to use clever methods of ANDing always true statements or ORing ones that do not change the result"

    I don't think I'm grasping what exactly that means, ands thats probably why I'm not seeing how to work out the problem. More specifically, what would be an "always true statement"?
     
  9. Apr 1, 2008 #8
    Well you already used it once: xy=xy(c+c') = xyc + xyc'. Obviously c+c' is always 1, so I have not changed the logical evaluation of that particular part at all and yet I have more pieces of the puzzle to use. That one xy term turned into 2 different terms that you can combine with the other terms. Might I suggest that you combine x'yc + xyc to form (x'+x)yc = yc.

    Can you do something similar with the rest using the other technique I suggested? Remember, X+X=X... or more accurately X = X + X. Maybe you find one term would be useful in simplifying two others?
     
    Last edited: Apr 1, 2008
  10. Apr 1, 2008 #9
    I tried this, but I don't believe it was right.

    x'c(y+y)+y'c(x+x)+xy(c+c')

    what is the correct way to use x=x+x?
     
  11. Apr 1, 2008 #10
    Ok well proving -> is proving hard for you. Perhaps you should try to prove it by starting with the right side and transforming it into the part that's left of the equals sign. Try this: Make each term on the right side of the equals sign a function of all of x, y and c. For example, yc=yc(x+x'). Distribute the terms and see if you can't make some of the distributed terms disappear. Be conscious at all times of what you want! Showing how the right side of the equation transforms into the left is a slightly simpler process.
     
    Last edited: Apr 1, 2008
  12. Apr 1, 2008 #11
    going from the right side I get this:

    yc(x+x')+xc(y+y')+xy(c+c') >>> xyc + x'yc +xyc+xy'c+xyc+xyc'

    the three xyc terms should reduce down to just one xyc using this:

    xy(c+c)

    so I get: x'yc+xy'c+xyc+xyc' >>> x'yc+xy'c+xy(c+c') >>> x'yc+xy'c+xy

    would that be correct?
     
  13. Apr 1, 2008 #12
    bingo
     
  14. Apr 1, 2008 #13
    and so to show from the left to the right, it's just the reverse of what I just did, right? Anyways, thanks for the help.
     
  15. Apr 1, 2008 #14
    Yeah pretty much, no problemo
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Boolean algebra
  1. Expand boolean algebra (Replies: 1)

  2. Boolean algebra (Replies: 2)

Loading...