Results 1 to 8 of 8

Math Help - intuitionalistic proofs

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    73

    intuitionalistic proofs

    I need some help deciding if these inferences are intuitionalistically valid

    1) Premise: There exists an x Fx----> for all x Gx
    Conclusion: for all x(Fx--->Gx)

    2) Premise: There exists an x(Fx--->Gx)
    Conclusion:for all x Fx---->there exists an x Gx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Maybe you should stick to scuba.

    So far, you're making no mathematicalogical sensicality.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    73
    well those are questions out of a logic text book so ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,513
    Thanks
    769
    These are perfectly valid questions about intuitionistic, or constructive, logic.

    Scubasteve123, when you are asking questions with narrow specialization, it may make sense to briefly describe the context; otherwise, people not familiar with the subject don't know whether it is nonsense or just something they don't know.

    1) Premise: There exists an x Fx----> for all x Gx
    Conclusion: for all x(Fx--->Gx)
    I assume parentheses are as follows: (\exists x\,F x)\to\forall x\,Gx implies \forall x\,(Fx\to Gx). This is true. Fix an x and assume F x. From this we get \exists y\,F y, and, therefore, \forall y\,Gy and so G x.

    2) Premise: There exists an x(Fx--->Gx)
    Conclusion:for all x Fx---->there exists an x Gx
    The conclusion is (\forall x\,Fx)\to\exists x\,Gx, right? This is also true. Assume \forall x\,Fx. From the assumption \exists x\,(Fx\to Gx) take an x that makes Fx\to Gx true. Then Fx is true, so Gx is true as well. Therefore, \exists x\,Gx.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    73
    thank you, all the symbols can be quite confusing, I apologise for the confusion I will be more clear in the future.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2009
    Posts
    73
    im having one hopefully last issue.
    in the text it seems to say that in intuitionist logic you can take negations away but cannot add the negations

    ie: you can get a proof of p from a proof of <not><not>p (double negation p)

    but i am under the impression that you cannot add negations, so one of the questions is to show you can conclude <not> <not> p from a proof of p.

    I didnt think this was valid?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,513
    Thanks
    769
    ie: you can get a proof of p from a proof of <not><not>p (double negation p)
    It's the other way around. From p one can prove (not (not p)) but in general not conversely.

    Usually, \neg p is taken as an abbreviation for p\to\bot ( \bot denotes falsehood). So, given p and an assumption p\to\bot one gets \bot by Modus Ponens. Closing the assumption p\to\bot one gets (p\to\bot)\to\bot, i.e., \neg\neg p.

    An intriguing thing is that one does not use the properties of \bot in the derivation above. I.e., from p one can prove (p\to q)\to q for any formula q.

    One can also remove double negations, but only if the remaining formula itself starts with a negation: \neg\neg\neg p\to\neg p. In general, there is no difference between classical and intuitionistic logic for formulas that start with a negation: \vdash\neg p classically implies \ vdash\neg p intuitionistically (Glivenko's theorem).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2009
    Posts
    73
    ive got it now thank you for the insight
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. i.d. proofs
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 20th 2010, 05:17 AM
  2. Proofs
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 17th 2008, 04:39 PM
  3. Proofs
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 17th 2008, 03:18 PM
  4. Set Proofs
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 5th 2008, 05:31 PM
  5. Replies: 3
    Last Post: October 6th 2007, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum