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Prove this math logic

  1. Oct 31, 2006 #1
    I need to prove the following:

    [tex]\forall x\forall y[(fy=x)\rightarrow Qx]\vdash \exists xQx[/tex].

    I can't do it.
     
  2. jcsd
  3. Oct 31, 2006 #2
    Is this even true? Seems a bit fishy.
     
  4. Oct 31, 2006 #3

    0rthodontist

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    I take it f is a function. It shouldn't seem fishy--after all fy has to equal something (most logics contain the assumption that the universe contains at least one object), and whatever it equals must have property Q.

    I am not familiar with the rules of your specific inference system. One plausible way, whether it's sufficiently formal for a logic derivation I do not know, is to instantiate the premise with y and fy, so you get
    (fy = fy) -> Qfy
    And maybe you have a premise that x = x for all x, so you could then derive Qfy
    Then use existential generalization on fy (is this legal in your system?) to obtain
    [tex]\exists[/tex]x Qx
     
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