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Biconditional equivalences

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    I have to prove that !(p <-> q) = (p <-> !q)


    2. The attempt at a solution
    I started by trying to just work out what each side of the equation was. So, starting with the left hand side
    !(p <-> q) = !((p->q) * (q->p)) (biconditional law)
    = !((!p + q) * (!q + p)) (implication law)
    = !(!p + q) + !(!q + p) (DeMorgan's theorem)
    = (p * !q) + (q * !p) (DeMorgan's theorem)

    Then, for the right hand side, I tried doing similarly:
    (p <-> !q) = (p -> !q) * (!q -> p) (biconditional law)
    = (!p + !q) * (q + p) (implication law)

    But, here I get stuck... I've thought about applying DeMorgan's, but it seems like the two expressions won't be equivalent if I do this. I've verified this equivalence is true using truth tables, but the equivalence logic seems to be escaping me. Can anyone lend a hand?

    Thanks! :)

    -Max
     
  2. jcsd
  3. Feb 1, 2010 #2
    A truth table would be much faster, but if you don't want that, you-ll have to develop (!p + !q) * (q + p), using the distributivity of *.
     
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