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Using existential generalization

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  1. Feb 10, 2017 #1
    1. The problem statement, all variables and given/known data
    is my method valid?

    ∃x¬R(x,a) --> ¬∃R(a,x)
    ¬R(a,a)
    thus, ¬R(a,b)

    2. Relevant equations
    N/A

    3. The attempt at a solution
    ∃x¬R(x,a) by existential gen. of ¬R(a,a)
    ¬∃R(a,x) by modus ponens
    ∀x¬R(a,x) by identity of ¬∃R(a,x)
    ¬R(a,b) by universal instantiation of ∀x¬R(a,x)
     
  2. jcsd
  3. Feb 10, 2017 #2

    fresh_42

    Staff: Mentor

    I have trouble to understand your notation. ##\lnot R(x,a)## suggests, that it is a statement, whereas ##\exists R(a,x)## looks like an element somewhere. Furthermore ##a## isn't quantified.
     
  4. Feb 10, 2017 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    As fresh_42 pointed out, the usual notation would be something like ##\lnot( \exists x ( R(a,x) )##

    Your general approach is correct.
     
  5. Feb 10, 2017 #4
    thank you stephen!
     
  6. Feb 10, 2017 #5
    yes i missed an "x" there. how was "a" not quantified?
     
  7. Feb 10, 2017 #6

    fresh_42

    Staff: Mentor

    Well, is ##a## fixed? Then it could be taken as part of ##R##, if ##R## is symmetric, which I don't know. Or does it mean for all ##a##, or there is an ##a##? Since it is in all occurrences of ##R## I tend to put it into the definition of ##R## to get rid of it, as it seems to be unnecessary. But then there are ##R(a,x)## and ##R(x,a)## and I don't know. what is the difference between them.
     
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