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How to work that out?

  1. Mar 20, 2005 #1
    1) (C -> A) and (!C -> B)
    2) (A and C) or (!C and B)

    I use the logic calculator and found these two formulas are logical equivalent, but I spend 2 hours there and still can't prove it because I can't eliminate one of the three items.

    Can anyone help me?

    Thanks in advance!
     
    Last edited: Mar 20, 2005
  2. jcsd
  3. Mar 20, 2005 #2
    ((C -> A) & (~C -> B)) <-> ((A & C) v (~C v B)), is not valid.

    It fails if A=B=C=false.
     
  4. Mar 20, 2005 #3
    sorry, man, I made a mistake here,
    the second one should be:
    ((A & C) v (~C & B)), it should be equivalent.


     
  5. Mar 20, 2005 #4
    (C -> A) :: A v ~C
    (~C -> B) :: C v B
    So ((C -> A) & (~C -> B)) :: (A v ~C) & (C v B)
    :: ((A v ~C) & C) v ((A v ~C) & B)
    :: ((A & C) v (~C & C)) v ((A & B) v (~C & B))
    :: (A & C) v (A & B) v (~C & B)
    :: AC + AB + ~CB (change of notation)

    I expanded it out into canonical form and combined terms:

    :: AC(B + ~B) + AB(C + ~C) + ~CB(A + ~A)
    :: ACB + AC~B + ~CBA + ~CB~A + ABC + AB~C
    :: ABC + AC~B + ~CBA + ~CB~A
    :: ~CB + AC
     
    Last edited: Mar 20, 2005
  6. Mar 21, 2005 #5
    Thank you so much!

    U r so helpful, man. thanx!

     
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