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Re: [aima-talk] Re: Quantifying over functions in first-order logic

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  • Peter Norvig
    I agree that for most of what you would reasonably want to do, it wouldn t matter. But the point is that when you quantify over all functions/relations
    Message 1 of 9 , Jun 13, 2006
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      I agree that for most of what you would reasonably want to do, it
      wouldn't matter. But the point is that when you quantify over all
      functions/relations (represented as objects), you wouldn't be
      quantifying over "call_function" nor "reflexive".

      -Peter

      On 6/13/06, cedricstjl <cedricstjl@...> wrote:
      > Why does it matter? I'm quantifying over "objects that behave exactly
      > like FOL functions" instead of quantifying directly over FOL functions.
      >
      > I could write a program, which takes as input a second-order logic
      > sentence, does inference in FOL with the conversion from my first
      > post, and translates back into second-order logic. Or are there SOL
      > sentences that wouldn't be translatable?
      >
      > Thank you for your time,
      >
      > Cedric
      >
      > --- In aima-talk@yahoogroups.com, "Peter Norvig" <peter@...> wrote:
      > >
      > > You could do that, and that could be a way of expressing reflexivity,
      > > but it wouldn't be quantifying over FOL functions -- it would be
      > > quantifying over objects that you happen to call "functions" but are
      > > regular objects as far as the logic is concerned.
      > >
      > > -Peter
      > >
      > > On 6/8/06, cedricstjl <cedricstjl@...> wrote:
      > > > I have some difficulty seeing the difference in expressiveness between
      > > > first-order logic and higher-order logics. Wikipedia (as well as AIMA,
      > > > AFAICT) says that F-O logic cannot quantify over functions. But
      > > > couldn't it be achieved equivalently by reifying functions and
      > > > predicates? I.e.:
      > > >
      > > > x = fun(var)
      > > >
      > > > becomes
      > > >
      > > > x = call_function(fun, var)
      > > >
      > > > and then one can express reflexivity:
      > > >
      > > > for all fun, a, b: reflexive(fun) <=> call_function(fun, a, b) =
      > > > call_function(fun, b, a)
      > > >
      > > > Cedric
      >
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    • cedricstjl
      I don t see why one would ever want to quantify over call_function. I don t think that there is an equivalent to that in regular logic. I can quantify over
      Message 2 of 9 , Jun 14, 2006
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        I don't see why one would ever want to quantify over call_function. I
        don't think that there is an equivalent to that in "regular" logic.

        I can quantify over reflexive by adding another level of indirection.
        reflexive(f) becomes call_function_2(reflexive, f). As long as all
        functions have a clearly defined level, it should work. Furthermore,
        using levels is, from what I've read, a common way of avoiding
        Russell's paradox in SOL. Does the "SOL is strictly more expressive
        than FOL" claim apply only to SOL without levels?

        Cedric

        --- In aima-talk@yahoogroups.com, "Peter Norvig" <peter@...> wrote:
        >
        > I agree that for most of what you would reasonably want to do, it
        > wouldn't matter. But the point is that when you quantify over all
        > functions/relations (represented as objects), you wouldn't be
        > quantifying over "call_function" nor "reflexive".
        >
        > -Peter
        >
        > On 6/13/06, cedricstjl <cedricstjl@...> wrote:
        > > Why does it matter? I'm quantifying over "objects that behave exactly
        > > like FOL functions" instead of quantifying directly over FOL
        functions.
        > >
        > > I could write a program, which takes as input a second-order logic
        > > sentence, does inference in FOL with the conversion from my first
        > > post, and translates back into second-order logic. Or are there SOL
        > > sentences that wouldn't be translatable?
        > >
        > > Thank you for your time,
        > >
        > > Cedric
        > >
        > > --- In aima-talk@yahoogroups.com, "Peter Norvig" <peter@> wrote:
        > > >
        > > > You could do that, and that could be a way of expressing
        reflexivity,
        > > > but it wouldn't be quantifying over FOL functions -- it would be
        > > > quantifying over objects that you happen to call "functions" but are
        > > > regular objects as far as the logic is concerned.
        > > >
        > > > -Peter
        > > >
        > > > On 6/8/06, cedricstjl <cedricstjl@> wrote:
        > > > > I have some difficulty seeing the difference in expressiveness
        between
        > > > > first-order logic and higher-order logics. Wikipedia (as well
        as AIMA,
        > > > > AFAICT) says that F-O logic cannot quantify over functions. But
        > > > > couldn't it be achieved equivalently by reifying functions and
        > > > > predicates? I.e.:
        > > > >
        > > > > x = fun(var)
        > > > >
        > > > > becomes
        > > > >
        > > > > x = call_function(fun, var)
        > > > >
        > > > > and then one can express reflexivity:
        > > > >
        > > > > for all fun, a, b: reflexive(fun) <=> call_function(fun, a, b) =
        > > > > call_function(fun, b, a)
        > > > >
        > > > > Cedric
      • cedricstjl
        Actually, I think that I don t even need levels. for all fun, a, b: call_function(reflexive, fun) call_function(fun, a, b) = call_function(fun, b, a) seems
        Message 3 of 9 , Jun 14, 2006
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          Actually, I think that I don't even need levels.

          for all fun, a, b: call_function(reflexive, fun) <=>
          call_function(fun, a, b) = call_function(fun, b, a)

          seems to work fine to me.

          Cedric

          --- In aima-talk@yahoogroups.com, "cedricstjl" <cedricstjl@...> wrote:
          >
          > I don't see why one would ever want to quantify over call_function. I
          > don't think that there is an equivalent to that in "regular" logic.
          >
          > I can quantify over reflexive by adding another level of indirection.
          > reflexive(f) becomes call_function_2(reflexive, f). As long as all
          > functions have a clearly defined level, it should work. Furthermore,
          > using levels is, from what I've read, a common way of avoiding
          > Russell's paradox in SOL. Does the "SOL is strictly more expressive
          > than FOL" claim apply only to SOL without levels?
          >
          > Cedric
          >
          > --- In aima-talk@yahoogroups.com, "Peter Norvig" <peter@> wrote:
          > >
          > > I agree that for most of what you would reasonably want to do, it
          > > wouldn't matter. But the point is that when you quantify over all
          > > functions/relations (represented as objects), you wouldn't be
          > > quantifying over "call_function" nor "reflexive".
          > >
          > > -Peter
          > >
          > > On 6/13/06, cedricstjl <cedricstjl@> wrote:
          > > > Why does it matter? I'm quantifying over "objects that behave
          exactly
          > > > like FOL functions" instead of quantifying directly over FOL
          > functions.
          > > >
          > > > I could write a program, which takes as input a second-order logic
          > > > sentence, does inference in FOL with the conversion from my first
          > > > post, and translates back into second-order logic. Or are there SOL
          > > > sentences that wouldn't be translatable?
          > > >
          > > > Thank you for your time,
          > > >
          > > > Cedric
          > > >
          > > > --- In aima-talk@yahoogroups.com, "Peter Norvig" <peter@> wrote:
          > > > >
          > > > > You could do that, and that could be a way of expressing
          > reflexivity,
          > > > > but it wouldn't be quantifying over FOL functions -- it would be
          > > > > quantifying over objects that you happen to call "functions"
          but are
          > > > > regular objects as far as the logic is concerned.
          > > > >
          > > > > -Peter
          > > > >
          > > > > On 6/8/06, cedricstjl <cedricstjl@> wrote:
          > > > > > I have some difficulty seeing the difference in expressiveness
          > between
          > > > > > first-order logic and higher-order logics. Wikipedia (as well
          > as AIMA,
          > > > > > AFAICT) says that F-O logic cannot quantify over functions. But
          > > > > > couldn't it be achieved equivalently by reifying functions and
          > > > > > predicates? I.e.:
          > > > > >
          > > > > > x = fun(var)
          > > > > >
          > > > > > becomes
          > > > > >
          > > > > > x = call_function(fun, var)
          > > > > >
          > > > > > and then one can express reflexivity:
          > > > > >
          > > > > > for all fun, a, b: reflexive(fun) <=> call_function(fun, a, b) =
          > > > > > call_function(fun, b, a)
          > > > > >
          > > > > > Cedric
          >
        • Joe Hendrix
          ... This is a useful technique. I ve also seen it called currifying in other papers, however it doesn t let you achieve the full power of second-order logic,
          Message 4 of 9 , Jun 14, 2006
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            > On 6/8/06, cedricstjl <cedricstjl@> wrote:
            > couldn't it be achieved equivalently by reifying functions and
            > predicates?

            This is a useful technique. I've also seen it called currifying in
            other papers, however it doesn't let you achieve the full power of
            second-order logic, much less higher-order logic.

            A basic first-order logic result is that finiteness is not
            expressible. That is there is no sentence P such that M is a model of
            P iff. M is finite. However, in second order logic one can express
            finiteness. Specifically, if P is the statement that all injective
            functions are surjective, then M is a model of P iff. M is finite.

            The wikipedia article on second-order logic discusses some of the
            differences between first-order and second-order logic at more length
            than this email.

            Joe
          • cedricstjl
            Hmmm, I ve got a hard time translating your example, or those from the Wikipedia article into logical sentences (I have no idea how to express that a model is
            Message 5 of 9 , Jun 15, 2006
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              Hmmm, I've got a hard time translating your example, or those from the
              Wikipedia article into logical sentences (I have no idea how to
              express that a model is finite, for instance). I guess I should pick
              up a book on the subject; my only background in logic is from AIMA and
              Hofstadter's book.

              Thanks for the info!

              Cedric

              --- In aima-talk@yahoogroups.com, Joe Hendrix <jhendrix@...> wrote:
              >
              > > On 6/8/06, cedricstjl <cedricstjl@> wrote:
              > > couldn't it be achieved equivalently by reifying functions and
              > > predicates?
              >
              > This is a useful technique. I've also seen it called currifying in
              > other papers, however it doesn't let you achieve the full power of
              > second-order logic, much less higher-order logic.
              >
              > A basic first-order logic result is that finiteness is not
              > expressible. That is there is no sentence P such that M is a model of
              > P iff. M is finite. However, in second order logic one can express
              > finiteness. Specifically, if P is the statement that all injective
              > functions are surjective, then M is a model of P iff. M is finite.
              >
              > The wikipedia article on second-order logic discusses some of the
              > differences between first-order and second-order logic at more length
              > than this email.
              >
              > Joe
              >
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