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Math Help - Herbrand Logic

  1. #1
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    Herbrand Logic

    Given \forall x.(p(x) \implies q(x))\quad and \quad p(a) use the Fitch system to prove q(a)


    I have started:



    1) \forall X.(p(X) \implies q(X)) \qquad (Premise)\\2) p(a) \qquad (Premise)\\3) \exists X.p(X) \qquad (Existential \quad Introduction:2)



    but I do not know how to continue. I believe I must prove also \forall x.(p(x) \implies q(a)) and by existential elimination I will get q(a). Can you help me solve it?
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  2. #2
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    Re: Herbrand Logic

    This is much simpler. Instantiate x with a in ∀x (p(x) => q(x)) and then use Modus Ponens (implication elimination) with p(a).
    Thanks from ManosG
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  3. #3
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    Re: Herbrand Logic

    Thanks!
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