Shannons Expansion Help

Discussion in 'Homework Help' started by knmonster, Oct 3, 2012.

  1. knmonster

    Thread Starter New Member

    Jun 8, 2012
    6
    0
    I have to prove
    ab + b'cd + acd = ab + b'cd
    I figured it out using purely Boolean algebra but I also have to prove it by using Shannon's Expansion. But I'm not really sure how to use Shannons Expansion and then notes aren't very helpful.
    I chose a, but I don't know how to choose the best term.

    This is the best I could figure out:
    f(a,b,c,d) = ab+ b'cd + acd = af(1,b,c,d) + a'f(0,b,c,d)
    f(1,b,c,d) = 1b + b'cd +1cd
    = b + b'cd + cd
    = b + cd (b+1)
    = b + cd

    f(0,b,c,d) = 0b + b'cd + 0cd
    = b'cd

    f(a,b,c,d) = a(b + cd) + a'(b'cd)
    = ab + acd +a'b'cd

    Have I at least started this right? I don't know where to go from here.
    Thanks
     
  2. knmonster

    Thread Starter New Member

    Jun 8, 2012
    6
    0
    whoops I think I figured it out:
    f(a,b,c,d) = a(b + cd) + a'(b'cd)
    = ab + acd +a'b'cd
    = ab+ abcd + ab'cd + a'b'cd (combining)
    = ab * 1 + abcd + ab'cd + a'b'cd (identity)
    = ab (1 + cd) + b'cd (a + a') (distributivity?) forgot which rule
    = ab (1) + b'cd (1) (Null element and complement)
    = ab + b'cd

    Is that correct?

    Also for the step
    = ab+ abcd + ab'cd + a'b'cd (combining)
    can i just go to
    = ab + b'cd using the covering theorem?

    Thanks for all the help
     
Loading...