Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

First order proof

  1. Mar 13, 2005 #1
    To do a tableau proof of this statement:
    [tex](\forall x) [P(x) \vee Q(x)] \supset [(\exists x)(P(x) \vee (\exists x)(Q(x)] [/tex]
    I started out by restating it as follows:
    [tex](\forall x) [P(x) \vee Q(x)] \supset [(\exists y)(P(y) \vee (\exists z)(Q(z)] [/tex]
    to avoid confusion over what's bound to what (and when).

    Is my approach:
    valid?
    invalid?
    recommended?
    not?
    a good idea?
    not?
    some other adjective (or expletive)?
     
  2. jcsd
  3. Mar 14, 2005 #2
    Your approach is correct.

    1. Ey(P(y) v EzQ(z)) <-> (EyP(y) v EzQ(z)),

    by, Ey(P(y) v p) <-> (EyP(y) v p).

    2. (EyP(y) v EzQ(z)) <-> (ExP(x) v ExQ(x)),

    by: EyP(y) <-> ExP(x), EzQ(z) <-> ExQ(x).

    3. Ey(P(y) v EzQ(z)) <-> (ExP(x) v ExQ(x)),

    by:1, 2, ((p <-> q) & (q <-> r)) -> (p <-> r).
     
  4. Mar 14, 2005 #3
    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: First order proof
  1. First order logic (Replies: 3)

  2. First order Logic (Replies: 6)

  3. First order logic (Replies: 0)

Loading...