Mixing
Set of primitive recursive functions is not finitely generated
$begingroup$
Let $PR$ be the set of of [primitive recursive functions][1]. Given $Xsubset PR$, let
$$
hat X={ h(g_1(vec x),dots,g_k(vec x)):kin mathbb{N}wedge h,g_1,dots,g_kin X},
$$
i.e. $hat X$ is obtained from $X$ by taking compositions.
Fix $Fsubset PR$ finite and let $I$ be the set of initial functions, i.e. $I$ consists of the unary successor function, the unary constantly $0$ function and all the projections. Define:
$$A_0=Fcup I$$
$$A_{n+1}=hat{A_n}$$
$$A=bigcup_{ninmathbb{N}}A_n$$
I'd like to show that $Aneq PR$.
In other words, the set of primitive recursive functions cannot be obtained from finitely many functions (plus the initial functions) just by taking compositions, i.e. the recursion part of definition is really necessary to get all the primitive recursive functions.
This is clearly not true if $F$ is not finite and just countable, since then we can just take $F=PR$, so $F$ being finite should play an important role.
Any hints?
EDIT: I've changed the question since the previous one was not very well formulated.
logic recursion
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
$endgroup$
add a comment |
$begingroup$
Let $PR$ be the set of of [primitive recursive functions][1]. Given $Xsubset PR$, let
$$
hat X={ h(g_1(vec x),dots,g_k(vec x)):kin mathbb{N}wedge h,g_1,dots,g_kin X},
$$
i.e. $hat X$ is obtained from $X$ by taking compositions.
Fix $Fsubset PR$ finite and let $I$ be the set of initial functions, i.e. $I$ consists of the unary successor function, the unary constantly $0$ function and all the projections. Define:
$$A_0=Fcup I$$
$$A_{n+1}=hat{A_n}$$
$$A=bigcup_{ninmathbb{N}}A_n$$
I'd like to show that $Aneq PR$.
In other words, the set of primitive recursive functions cannot be obtained from finitely many functions (plus the initial functions) just by taking compositions, i.e. the recursion part of definition is really necessary to get all the primitive recursive functions.
This is clearly not true if $F$ is not finite and just countable, since then we can just take $F=PR$, so $F$ being finite should play an important role.
Any hints?
EDIT: I've changed the question since the previous one was not very well formulated.
logic recursion
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
$endgroup$
$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
1
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26
add a comment |
$begingroup$
Let $PR$ be the set of of [primitive recursive functions][1]. Given $Xsubset PR$, let
$$
hat X={ h(g_1(vec x),dots,g_k(vec x)):kin mathbb{N}wedge h,g_1,dots,g_kin X},
$$
i.e. $hat X$ is obtained from $X$ by taking compositions.
Fix $Fsubset PR$ finite and let $I$ be the set of initial functions, i.e. $I$ consists of the unary successor function, the unary constantly $0$ function and all the projections. Define:
$$A_0=Fcup I$$
$$A_{n+1}=hat{A_n}$$
$$A=bigcup_{ninmathbb{N}}A_n$$
I'd like to show that $Aneq PR$.
In other words, the set of primitive recursive functions cannot be obtained from finitely many functions (plus the initial functions) just by taking compositions, i.e. the recursion part of definition is really necessary to get all the primitive recursive functions.
This is clearly not true if $F$ is not finite and just countable, since then we can just take $F=PR$, so $F$ being finite should play an important role.
Any hints?
EDIT: I've changed the question since the previous one was not very well formulated.
logic recursion
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
$endgroup$
Let $PR$ be the set of of [primitive recursive functions][1]. Given $Xsubset PR$, let
$$
hat X={ h(g_1(vec x),dots,g_k(vec x)):kin mathbb{N}wedge h,g_1,dots,g_kin X},
$$
i.e. $hat X$ is obtained from $X$ by taking compositions.
Fix $Fsubset PR$ finite and let $I$ be the set of initial functions, i.e. $I$ consists of the unary successor function, the unary constantly $0$ function and all the projections. Define:
$$A_0=Fcup I$$
$$A_{n+1}=hat{A_n}$$
$$A=bigcup_{ninmathbb{N}}A_n$$
I'd like to show that $Aneq PR$.
In other words, the set of primitive recursive functions cannot be obtained from finitely many functions (plus the initial functions) just by taking compositions, i.e. the recursion part of definition is really necessary to get all the primitive recursive functions.
This is clearly not true if $F$ is not finite and just countable, since then we can just take $F=PR$, so $F$ being finite should play an important role.
Any hints?
EDIT: I've changed the question since the previous one was not very well formulated.
logic recursion
logic recursion
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
edited Feb 4 at 2:35
Reveillark
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
asked Feb 1 at 4:36
ReveillarkReveillark
4,801822
4,801822
$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
1
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26
add a comment |
$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
1
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26
$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
1
1
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26
add a comment |
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$begingroup$
You are using $n$ twice, I think for two different things. Are you assuming your functions $g$ in PR all have the form $g:mathbb{R}^nrightarrow mathbb{R}$, for some fixed positive integer $n$? It may also help to further illuminate/define the set PR.
$endgroup$
– Michael
Feb 1 at 5:31
$begingroup$
@Michael There´s no constraint on the arity of primitive recursive functions, which go from $mathbb{N}^n$ to $mathbb{N}$ for different $n$. I've changed some of the notation, perhaps it´s more clear now? Plus I linked the definition of primitive recursive.
$endgroup$
– Reveillark
Feb 1 at 21:29
1
$begingroup$
I may be being stupid, but don't the functions in your set $H$ only depend on at most a fixed prefix of the argument vector $vec{x}$? I.e., as $F$ is finite, there is an $n$ such that no function in $F$ and hence in $H$ depends on $x_i$ for $i > n$.
$endgroup$
– Rob Arthan
Feb 1 at 22:42
$begingroup$
@RobArthan I think you're right. I'll think about the problem some more to see if I can state it in a way that makes sense.
$endgroup$
– Reveillark
Feb 2 at 0:26