Does this formalism adequately describe functions of one variable?

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2












$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:45















2












$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:45













2












2








2





$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.










share|cite|improve this question











$endgroup$




Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.







functions elementary-set-theory logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 20 at 1:56







Dan Christensen

















asked Feb 20 at 1:15









Dan ChristensenDan Christensen

8,64321835




8,64321835











  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:45
















  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:45















$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24




$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24




2




2




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24












$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36




$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36












$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38




$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38












$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45




$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45










3 Answers
3






active

oldest

votes


















5












$begingroup$

For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






share|cite|improve this answer











$endgroup$




















    3












    $begingroup$

    Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
    beginalign*
    (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
    endalign*






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      I am familiar with the that particular formalism.Could you comment on the one I presented here?
      $endgroup$
      – Dan Christensen
      Feb 20 at 1:54











    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      Feb 20 at 1:57


















    0












    $begingroup$

    If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
      $endgroup$
      – Dan Christensen
      Feb 21 at 15:43










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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



    Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



    $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



    If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



    1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






    share|cite|improve this answer











    $endgroup$

















      5












      $begingroup$

      For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



      Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



      $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



      If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



      1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






      share|cite|improve this answer











      $endgroup$















        5












        5








        5





        $begingroup$

        For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



        Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



        $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



        If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



        1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






        share|cite|improve this answer











        $endgroup$



        For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



        Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



        $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



        If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



        1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 20 at 2:59

























        answered Feb 20 at 2:46









        Derek ElkinsDerek Elkins

        17.3k11437




        17.3k11437





















            3












            $begingroup$

            Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57















            3












            $begingroup$

            Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57













            3












            3








            3





            $begingroup$

            Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            share|cite|improve this answer









            $endgroup$



            Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 20 at 1:22









            APC89APC89

            2,341720




            2,341720











            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57
















            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57















            $begingroup$
            I am familiar with the that particular formalism.Could you comment on the one I presented here?
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:54





            $begingroup$
            I am familiar with the that particular formalism.Could you comment on the one I presented here?
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:54













            $begingroup$
            I guess the heading wasn't very clear. I have changed it.
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:57




            $begingroup$
            I guess the heading wasn't very clear. I have changed it.
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:57











            0












            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43















            0












            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43













            0












            0








            0





            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$



            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 20 at 9:07









            Alberto TakaseAlberto Takase

            2,380719




            2,380719











            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43
















            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43















            $begingroup$
            What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
            $endgroup$
            – Dan Christensen
            Feb 21 at 15:43




            $begingroup$
            What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
            $endgroup$
            – Dan Christensen
            Feb 21 at 15:43

















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