Troubles by understanding arity of terms in A Course in Universal Algebra.
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In A Course in Universal Algebra in definition 10.1 terms are introduced.
What puzzles me is the statement about arity:
"A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."
Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?
If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?
This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.
I really do not understand this definition.
If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?
soft-question definition universal-algebra
$endgroup$
add a comment |
$begingroup$
In A Course in Universal Algebra in definition 10.1 terms are introduced.
What puzzles me is the statement about arity:
"A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."
Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?
If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?
This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.
I really do not understand this definition.
If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?
soft-question definition universal-algebra
$endgroup$
add a comment |
$begingroup$
In A Course in Universal Algebra in definition 10.1 terms are introduced.
What puzzles me is the statement about arity:
"A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."
Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?
If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?
This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.
I really do not understand this definition.
If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?
soft-question definition universal-algebra
$endgroup$
In A Course in Universal Algebra in definition 10.1 terms are introduced.
What puzzles me is the statement about arity:
"A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."
Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?
If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?
This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.
I really do not understand this definition.
If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?
soft-question definition universal-algebra
soft-question definition universal-algebra
edited Dec 26 '18 at 11:27
drhab
asked Dec 23 '18 at 12:14
drhabdrhab
99.3k544130
99.3k544130
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2 Answers
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$begingroup$
I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.
Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.
Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^Vrightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.
- Note that there's no requirement here that $V$ be finite!
In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).
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Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
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You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
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– drhab
Dec 23 '18 at 19:46
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@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
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– William DeMeo
Dec 23 '18 at 19:53
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@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
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– drhab
Dec 23 '18 at 19:59
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@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
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– William DeMeo
Dec 23 '18 at 20:03
|
show 3 more comments
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Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:
Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:
- every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,
- if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.
Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:
- if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.
- if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.
The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:
Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.
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Thank you. I will take a good look at this (and also the answer of Noah).
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– drhab
Dec 23 '18 at 17:34
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The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
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– drhab
Dec 23 '18 at 19:21
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@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
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– William DeMeo
Dec 23 '18 at 20:04
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If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
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– William DeMeo
Dec 23 '18 at 20:08
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Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
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– drhab
Dec 24 '18 at 6:07
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2 Answers
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$begingroup$
I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.
Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.
Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^Vrightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.
- Note that there's no requirement here that $V$ be finite!
In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).
$endgroup$
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
|
show 3 more comments
$begingroup$
I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.
Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.
Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^Vrightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.
- Note that there's no requirement here that $V$ be finite!
In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).
$endgroup$
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
|
show 3 more comments
$begingroup$
I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.
Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.
Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^Vrightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.
- Note that there's no requirement here that $V$ be finite!
In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).
$endgroup$
I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.
Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.
Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^Vrightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.
- Note that there's no requirement here that $V$ be finite!
In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).
edited Dec 23 '18 at 20:15
answered Dec 23 '18 at 16:55
Noah SchweberNoah Schweber
123k10149284
123k10149284
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
|
show 3 more comments
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
Thank you Noah. Light starts shining, but (as always) I need time to digest.
$endgroup$
– drhab
Dec 23 '18 at 17:26
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
You are saying that every (finite I guess) set $V$ induces together with $p$ a map $p_V:A^{|V|}to A$. I understand, but shouldn't we say that for every linear ordered finite set $V$? If for instance $p$ is variable $x$ then it must be clear which projection (there are $|V|$ in total) is denoted by $p_V$, right? I asked a similar question to @Eran.
$endgroup$
– drhab
Dec 23 '18 at 19:46
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@drhab Yes, what you're saying is correct for finitary operations. However, a more general definition would be to interpret to domain $A^V$ as the collection of all functions from $V$ to $A$. Think of these elements of the domain as "$V$-tuples." Then, I would call $f : A^V to A$ a $V$-ary operation on $A$.
$endgroup$
– William DeMeo
Dec 23 '18 at 19:53
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@WilliamDeMeo Ah, so actually everything will be okay immediately if we replace $|V|$ by $V$ here so that $p_V:A^Vto A$. Then the order I was asking for is present. Next to that it has the advantage that it eventually can be expanded to infinite $V$. Do I understand you correct?
$endgroup$
– drhab
Dec 23 '18 at 19:59
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
$begingroup$
@drhab Yes, exactly! And see the comment I'm about to post after Erin's answer.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:03
|
show 3 more comments
$begingroup$
Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:
Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:
- every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,
- if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.
Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:
- if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.
- if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.
The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:
Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.
$endgroup$
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
add a comment |
$begingroup$
Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:
Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:
- every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,
- if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.
Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:
- if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.
- if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.
The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:
Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.
$endgroup$
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
add a comment |
$begingroup$
Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:
Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:
- every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,
- if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.
Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:
- if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.
- if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.
The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:
Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.
$endgroup$
Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:
Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:
- every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,
- if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.
Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:
- if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.
- if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.
The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:
Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.
answered Dec 23 '18 at 16:51
EranEran
1,260818
1,260818
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
add a comment |
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
Thank you. I will take a good look at this (and also the answer of Noah).
$endgroup$
– drhab
Dec 23 '18 at 17:34
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
The first 2. in your answer starts with "if $t_1,dots,t_k$.." but I suspect the $k$ must be $n$ here. Correct? Secondly after the second 1. in your answer you define $t^A$ as the $i$-th projection if $t=x_i$. So the index $i$ is determining. Doesn't that ask for a tuple $<x_1,dots,x_n>$ instead of a set $X_k={x_1,dots,x_n}$, or otherwise for a set $X_k$ having $k$ elements that is accompanied by a linear order? Maybe that was done implicitly already by introducing indices for the elements of $X_k$.
$endgroup$
– drhab
Dec 23 '18 at 19:21
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
@drhab When trying to write programs to handle general operations and terms, (Andrej Bauer showed me) it's easier to handle arities in a more general way. For finitary operations or terms, input arguments are finite tuples, like $a = (a_1, dots, a_n)$, but a tuple is really a function $a : {1,2,dots, n} to A$ (or replace $A$ with $T$ if you want input arguments to be terms). More generally, if $rho : F to alpha$ is a function from operation symbols to the "arity" type $alpha$, then operations have type $f : A^{rho f} to A$. The arity type $alpha$ can be fairly general.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:04
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
If you know Coq, and you want to see how this is actually implemented (in dependent type theory), look here, or if you know Lean, look here.
$endgroup$
– William DeMeo
Dec 23 '18 at 20:08
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
$begingroup$
Everything is clear to me now, and this answer was very very useful. So was the one of Noah (and also the comments of @WilliamDeMeo). Actually I regret that I can only accept one answer. Thanks a lot!
$endgroup$
– drhab
Dec 24 '18 at 6:07
add a comment |
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