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Proof in predicate calculus

  1. Jun 22, 2012 #1
    Given :


    1) c is a constant

    2) P and K are one place operation symbols

    3) G and H are a two place predicate symbols


    The following hypothesis

    1)for all A { G(A,A) }

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

    Then prove :

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

    2) for all A { H(A,c) => H( P(A),c) }
  2. jcsd
  3. Jun 23, 2012 #2
    How you formally work this through depends on what system you're asked to work with, but a strategy would be to let B=P(A).
  4. Jun 23, 2012 #3
    Thanks dcpo .So i put B=P(A) in 2 and i get :

    { H(A,c) =>( G[P(A),P(A)] <=> ( G[K(P(A)),A] and H(P(A),c)))}
    is that correct?

    The proof is in predicate calculus ,so every line of the proof has to be accounted for and justified
  5. Jun 23, 2012 #4
    Well, what you need to say depends on how formal you have to be, and on what deduction system you have to use, but what you have is the basis for a rigorous 'everyday' proof, so long as you make explicit the role of the [itex]\forall A(G(A,A))[/itex] hypothesis. If you haven't been given an explicit formal deduction system to work with that should be enough.
  6. Jun 23, 2012 #5
    The proof as i said is an ordinary proof in predicate calculus with the usal 4 general laws i.e

    1) Universal Elimination
    2) Universal Introduction
    3) Existential Elimination
    4) Existential Introduction

    Plus the rules of statement calculus
  7. Jul 7, 2012 #6
    Well, we ain't supposed to be doing homework problems here, but the proofs are extremely simple.

    for proof 1: Your notation is inconsistent, so can't help until you clean that up.

    for proof 2: [tex] \forall A (H(A,c) \Rightarrow H(P(A),c)) [/tex]
    Assuming A, H(A,c), and P(A) exist immediately gives result H(P(A),c) from b 1 and 2.
  8. Jul 14, 2012 #7
    Sorry to ask but do you know how a proof is done in predicate calculus?

    I mean have you done any predicate calculus?
  9. Jul 19, 2012 #8
    Um yeah ...
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