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Homework Help: Homework help: predicate logic

  1. May 4, 2005 #1
    I am taking a logic class and we are getting into Predicate Logic and i have no idea how to do it can someone help me?
  2. jcsd
  3. May 4, 2005 #2


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    What are some specific things you don't understand? Are you comfortable with everything in propositional (or sentential) logic?
  4. May 5, 2005 #3
    I'm also studying predicate logic, and I can't say I'm hugely comfortable with propositional logic.

    Here's a problem I'm working on right now:

    V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx)

    Having trouble getting started. Due in two hours. I know that I should hypothesize...

    Here's basically where I got stuck:

    1. V(universal quantifier)xFx |- VxGx--->Vx(Fx & Gx) A
    2. | VxGx H
    3. | Vx(Gx v Fx) 1, 2 vI
    4. VxGx ----> Vx(Gx v Fx) 2,3--->I

    From there I'm lost. De Morgan's doesn't get me exactly what I need.

    Eh...looks like I need to get out of the propositional logic box.
    Last edited: May 5, 2005
  5. May 5, 2005 #4
    Hell, all these are hard. The next ones are:

    3. -](makeshift particular quantifier)x]yLxy |- Vx-Lxx

    Quantifier exchange...but how do I get rid of the y?

    4. |- ]xFx v Vx-Fx

    Last edited: May 5, 2005
  6. May 5, 2005 #5


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    You can use Universal Instantiation, Conditional Proof, Conjunction Introduction (P, Q l- (P & Q)), and Universal Generalization.
    Is "-" negation or part of the quantifiers? "~" is negation, A and E are quantifiers.
    3) ~(ExEy(Lxy)) l- Ax~(Lxx) ?
    4) l- ~(Ex(Fx) v Ax~(Fx)) ?
  7. May 5, 2005 #6
    According to your definitions, those problems look like:

    3. ~Ex EyLxy |-Ax ~Lxx (close to the same thing)

    4.|- ExFx v Ax ~Fx (looks like my post had a typo - no negation on the first existential quantifier)

    Thanks, btw. Pretty sure I got the first one. I have:

    2. Fa AE
    3. | Ga
    4. | Ga & Fa
    5. Ga ---> Ga & Fa
    6. Ga ---> Fa & Ga
    7. AxGx--->Ax(Fx & Gx)

    Don't worry too much about it...need help on these others.
    Last edited: May 5, 2005
  8. May 5, 2005 #7
    Anyhow, thanks for the help. Might've gotten the last one, but I'm heading off to class now. Serves me right for procrastinating. :rolleyes:
  9. May 5, 2005 #8


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    Sorry, I'm not sure how to get rid of the y either.
    Double negation. What is ~~Ax(~Fx)? Or ~~Ex(Fx)?

    Looks good.

    Edit: Ah, I spent too long trying to find a stupid rule for 3. :grumpy:
    Last edited: May 5, 2005
  10. May 5, 2005 #9
    I did manage to get #4 right before I left for class. It's just:

    4.|- ExFx v Ax ~Fx

    1. | ~ExFx
    2. | Ax ~Fx 1, QE
    3. ~ExFx ---> Ax ~Fx 1-2 -->I
    4. ~~ExFx v Ax ~Fx 3, MI
    5. ExFx v Ax ~Fx

    That double negation might've been what you were hinting at. I was having trouble figuring out how Quantum Exchange worked exactly, heh. Oh well. I got another homework assignment...might end up posting it too. :tongue:
  11. May 5, 2005 #10


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    For #3, can't you use this?

    Ay Py

    Or some sort of substitution rule?
  12. May 5, 2005 #11


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    3. ~Ex EyLxy |-Ax ~Lxx
    If the negation applies only to Ex, I don't know what to do- it doesn't really even make sense to me. So I'll assume it applies to ExEy.
    I only know some predicate logic, and I'm least comfortable with quantifier inference rules, but I've read that you can't instantiate a negated quantifier (which makes sense); So you at least know you need to move the negation to Lxy beforehand: AxAy~(Lxy). Unless otherwise noted, x and y aren't necessarily distinct, so AxAy~(Lxy) implies Ax~(Lxx), but I don't know how to derive it. If there isn't a restriction on UI such that
    1. AxAy~(Lxy)
    2. ~Luu (u is an arbitrary constant) [1, UI]
    is invalid, the proof is a cinch. If that is invalid, I don't know what to do.
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