The set of all the indices of the members of a set












1












$begingroup$


Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, I am interested in the set of all the indices of the members of $X^{i}$, which in this case is ${i, j1, cdots, jn}$. Let $Q(i,j)$ be a binary relation. Then considering the free variable $i$, which one of the following formulas is syntactically and semantically valid and meaningful?



(1) $(forall j in [(arglimits_{k} x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(2) $(forall j in [(arg x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(3) $(forall j in [(arglimits_{k} forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(4) $(forall j in [(arg forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$






logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your square brackets notation mean? What is $arg$?
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 20:30










  • $begingroup$
    @RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
    $endgroup$
    – Pinton
    Dec 3 '18 at 22:37










  • $begingroup$
    This isn't standard mathematical notation. Please supply some more context.
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 23:16










  • $begingroup$
    @RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:40












  • $begingroup$
    @RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:45
















1












$begingroup$


Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, I am interested in the set of all the indices of the members of $X^{i}$, which in this case is ${i, j1, cdots, jn}$. Let $Q(i,j)$ be a binary relation. Then considering the free variable $i$, which one of the following formulas is syntactically and semantically valid and meaningful?



(1) $(forall j in [(arglimits_{k} x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(2) $(forall j in [(arg x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(3) $(forall j in [(arglimits_{k} forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(4) $(forall j in [(arg forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$






logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your square brackets notation mean? What is $arg$?
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 20:30










  • $begingroup$
    @RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
    $endgroup$
    – Pinton
    Dec 3 '18 at 22:37










  • $begingroup$
    This isn't standard mathematical notation. Please supply some more context.
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 23:16










  • $begingroup$
    @RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:40












  • $begingroup$
    @RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:45














1












1








1


1



$begingroup$


Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, I am interested in the set of all the indices of the members of $X^{i}$, which in this case is ${i, j1, cdots, jn}$. Let $Q(i,j)$ be a binary relation. Then considering the free variable $i$, which one of the following formulas is syntactically and semantically valid and meaningful?



(1) $(forall j in [(arglimits_{k} x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(2) $(forall j in [(arg x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(3) $(forall j in [(arglimits_{k} forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(4) $(forall j in [(arg forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$






logic







share|cite|improve this question











$endgroup$




Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, I am interested in the set of all the indices of the members of $X^{i}$, which in this case is ${i, j1, cdots, jn}$. Let $Q(i,j)$ be a binary relation. Then considering the free variable $i$, which one of the following formulas is syntactically and semantically valid and meaningful?



(1) $(forall j in [(arglimits_{k} x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(2) $(forall j in [(arg x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(3) $(forall j in [(arglimits_{k} forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$



(4) $(forall j in [(arg forall x^{k} in X^{i})setminus{i}])~~~Q(i,j)$






logic




logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 19:04







Pinton

















asked Dec 3 '18 at 17:57









PintonPinton

133




133












  • $begingroup$
    What does your square brackets notation mean? What is $arg$?
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 20:30










  • $begingroup$
    @RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
    $endgroup$
    – Pinton
    Dec 3 '18 at 22:37










  • $begingroup$
    This isn't standard mathematical notation. Please supply some more context.
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 23:16










  • $begingroup$
    @RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:40












  • $begingroup$
    @RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:45


















  • $begingroup$
    What does your square brackets notation mean? What is $arg$?
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 20:30










  • $begingroup$
    @RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
    $endgroup$
    – Pinton
    Dec 3 '18 at 22:37










  • $begingroup$
    This isn't standard mathematical notation. Please supply some more context.
    $endgroup$
    – Rob Arthan
    Dec 3 '18 at 23:16










  • $begingroup$
    @RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:40












  • $begingroup$
    @RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
    $endgroup$
    – Pinton
    Dec 3 '18 at 23:45
















$begingroup$
What does your square brackets notation mean? What is $arg$?
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:30




$begingroup$
What does your square brackets notation mean? What is $arg$?
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:30












$begingroup$
@RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
$endgroup$
– Pinton
Dec 3 '18 at 22:37




$begingroup$
@RobArthan: The brackets are just used to indicate the precedence of the applied operators. So, one can simply replace them by parenthesis or curly braces. Furthermore, arg is supposed to return the indice(s) of its argument.
$endgroup$
– Pinton
Dec 3 '18 at 22:37












$begingroup$
This isn't standard mathematical notation. Please supply some more context.
$endgroup$
– Rob Arthan
Dec 3 '18 at 23:16




$begingroup$
This isn't standard mathematical notation. Please supply some more context.
$endgroup$
– Rob Arthan
Dec 3 '18 at 23:16












$begingroup$
@RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
$endgroup$
– Pinton
Dec 3 '18 at 23:40






$begingroup$
@RobArthan: Sorry, but I don't know what you mean by more context. As I noted in the question, all I need is the correct notation to iterate over the set of all the indices of the members of a specific set. In particular arg works like argmax, except the former returns all arguments, but the latter reflects the arguments of the maximum value of a set.
$endgroup$
– Pinton
Dec 3 '18 at 23:40














$begingroup$
@RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
$endgroup$
– Pinton
Dec 3 '18 at 23:45




$begingroup$
@RobArthan: In other words, I intend to mathematically define the index set of a set. By definition, the index set includes the indices of the members of that set.
$endgroup$
– Pinton
Dec 3 '18 at 23:45










1 Answer
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$begingroup$

This way of combining arg with quantifiers looks a bit awkward. Instead, I would define an index set as follows.




Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, $I_{X}$ is the index
set of $X^{i}$ denoting the set of all the indices of the
elements of $X^{i}$ as follows begin{equation}
I_{X^{i}} := {arglimits_{k} x^{k} | forall
x^{k} in X^{i}}, end{equation} where $arg(cdot)$
operator returns the index of its arguments.




Now, one can simply say



$(forall j in I_{X^{i}})~~~ Q(i,j)$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

    oldest

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    0












    $begingroup$

    This way of combining arg with quantifiers looks a bit awkward. Instead, I would define an index set as follows.




    Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, $I_{X}$ is the index
    set     of $X^{i}$ denoting the set of all the indices of the
    elements of $X^{i}$ as follows begin{equation}
    I_{X^{i}} := {arglimits_{k} x^{k} | forall
    x^{k} in X^{i}}, end{equation}     where $arg(cdot)$
    operator returns the index of its arguments.




    Now, one can simply say



    $(forall j in I_{X^{i}})~~~ Q(i,j)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This way of combining arg with quantifiers looks a bit awkward. Instead, I would define an index set as follows.




      Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, $I_{X}$ is the index
      set       of $X^{i}$ denoting the set of all the indices of the
      elements of $X^{i}$ as follows begin{equation}
      I_{X^{i}} := {arglimits_{k} x^{k} | forall
      x^{k} in X^{i}}, end{equation}       where $arg(cdot)$
      operator returns the index of its arguments.




      Now, one can simply say



      $(forall j in I_{X^{i}})~~~ Q(i,j)$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This way of combining arg with quantifiers looks a bit awkward. Instead, I would define an index set as follows.




        Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, $I_{X}$ is the index
        set         of $X^{i}$ denoting the set of all the indices of the
        elements of $X^{i}$ as follows begin{equation}
        I_{X^{i}} := {arglimits_{k} x^{k} | forall
        x^{k} in X^{i}}, end{equation}         where $arg(cdot)$
        operator returns the index of its arguments.




        Now, one can simply say



        $(forall j in I_{X^{i}})~~~ Q(i,j)$






        share|cite|improve this answer









        $endgroup$



        This way of combining arg with quantifiers looks a bit awkward. Instead, I would define an index set as follows.




        Given $X^{i} := {x^{i}, x^{j1}, cdots, x^{jn}}$, $I_{X}$ is the index
        set         of $X^{i}$ denoting the set of all the indices of the
        elements of $X^{i}$ as follows begin{equation}
        I_{X^{i}} := {arglimits_{k} x^{k} | forall
        x^{k} in X^{i}}, end{equation}         where $arg(cdot)$
        operator returns the index of its arguments.




        Now, one can simply say



        $(forall j in I_{X^{i}})~~~ Q(i,j)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 18:03









        RoboticistRoboticist

        234315




        234315






























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