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Proof in predicate calculus

  1. Oct 31, 2011 #1
    Let:

    1)P be one place operation

    2)K be one place operation

    3) c be a constant


    let :

    1) G be a two place predicate

    2) H be a two place predicate

    Let :

    The following axioms or assumptions)

    1)for all A { H(A,c)v H(c,A)v G(A,c)}

    2)for all A { H(A,c)=> G[P(A),A]}

    3)for all A {H(c,A) => G[P(A),K(A)]}

    4)for all A {G[K(A),c] => G(A,c)}.

    5)for all A,B,C { [G(A,B) and G(A,C)]=> G(B,C)}

    Then formally prove :

    for all A {G[P(A),c] => G(A,c)}
     
  2. jcsd
  3. Oct 31, 2011 #2

    CompuChip

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    Sounds like an interesting exercise, I'll try it later.
    Have you finished it yet, or are you just fishing around for the answer? :-)
     
  4. Oct 31, 2011 #3

    CompuChip

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    Just got home from work and had a serious look at the problem.
    It's an interesting problem.
    I found the way to do it, but I haven't written down the complete formal proof because I don't really feel like proving OR elimination for three variables at this point (i.e. { A v B v C, A => D, B => D, C => D } |= D).
     
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