Prove that there is a $Delta_1$-set $E$ which satisfies $S_0 setminus S_1 subset E subset S_0$
$begingroup$
I'd appreciate some help for the following exercise. $Sigma_1, Pi_1$ and $Delta_1$ are defined here: https://en.wikipedia.org/wiki/Arithmetical_hierarchy
I think I can prove 2. when I have 1. - can someone give me hint, how to start with the proof? I already know that all $Sigma_1$-sets are r.e. and all $Pi_1$-sets are co-r.e. and all $Delta_1$-sets are recursive.
Assume $S_0, S_1 subset mathbb{N}$ are $Sigma_1$-sets which satisfy $S_0 cup S_1 = mathbb{N}$. Prove that there is a $Delta_1$-set $E$ such that $S_0 setminus S_1 subset E subset S_0$ is satisfied.
Assume $Q_0, Q_1 subset mathbb{N}$ are $Pi_1$-sets which satisfy $S_0 cap S_1 = emptyset$. Prove that there is a $Delta_1$-set $E$ such that $Q_0 subset E$ and $Q_1 subset mathbb{N} setminus E$ are satisfied.
EDIT:
I'm trying to construct $Pi_1$ and $Sigma_1$ formulas for $E$:
$exists k varphi_0(k,n),exists k varphi_1(k,n)$ are $Sigma_1$ formulas which define $S_0$ and $S_1$.
$S_0={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n)}$
$S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_1(k,n)}$
$S_0setminus S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land forall l ~ lnot varphi_1(l,n)}$
$E={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)}$
$exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)$ is equivalent to a $Sigma_1$ formula. Now I need a $Sigma_1$ formula for $mathbb{N} setminus E$ or a $Pi_1$ formula for $E$. Can someone give me a hint how to find it?
logic first-order-logic predicate-logic recursion computability
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
$endgroup$
add a comment |
$begingroup$
I'd appreciate some help for the following exercise. $Sigma_1, Pi_1$ and $Delta_1$ are defined here: https://en.wikipedia.org/wiki/Arithmetical_hierarchy
I think I can prove 2. when I have 1. - can someone give me hint, how to start with the proof? I already know that all $Sigma_1$-sets are r.e. and all $Pi_1$-sets are co-r.e. and all $Delta_1$-sets are recursive.
Assume $S_0, S_1 subset mathbb{N}$ are $Sigma_1$-sets which satisfy $S_0 cup S_1 = mathbb{N}$. Prove that there is a $Delta_1$-set $E$ such that $S_0 setminus S_1 subset E subset S_0$ is satisfied.
Assume $Q_0, Q_1 subset mathbb{N}$ are $Pi_1$-sets which satisfy $S_0 cap S_1 = emptyset$. Prove that there is a $Delta_1$-set $E$ such that $Q_0 subset E$ and $Q_1 subset mathbb{N} setminus E$ are satisfied.
EDIT:
I'm trying to construct $Pi_1$ and $Sigma_1$ formulas for $E$:
$exists k varphi_0(k,n),exists k varphi_1(k,n)$ are $Sigma_1$ formulas which define $S_0$ and $S_1$.
$S_0={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n)}$
$S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_1(k,n)}$
$S_0setminus S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land forall l ~ lnot varphi_1(l,n)}$
$E={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)}$
$exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)$ is equivalent to a $Sigma_1$ formula. Now I need a $Sigma_1$ formula for $mathbb{N} setminus E$ or a $Pi_1$ formula for $E$. Can someone give me a hint how to find it?
logic first-order-logic predicate-logic recursion computability
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
$endgroup$
add a comment |
$begingroup$
I'd appreciate some help for the following exercise. $Sigma_1, Pi_1$ and $Delta_1$ are defined here: https://en.wikipedia.org/wiki/Arithmetical_hierarchy
I think I can prove 2. when I have 1. - can someone give me hint, how to start with the proof? I already know that all $Sigma_1$-sets are r.e. and all $Pi_1$-sets are co-r.e. and all $Delta_1$-sets are recursive.
Assume $S_0, S_1 subset mathbb{N}$ are $Sigma_1$-sets which satisfy $S_0 cup S_1 = mathbb{N}$. Prove that there is a $Delta_1$-set $E$ such that $S_0 setminus S_1 subset E subset S_0$ is satisfied.
Assume $Q_0, Q_1 subset mathbb{N}$ are $Pi_1$-sets which satisfy $S_0 cap S_1 = emptyset$. Prove that there is a $Delta_1$-set $E$ such that $Q_0 subset E$ and $Q_1 subset mathbb{N} setminus E$ are satisfied.
EDIT:
I'm trying to construct $Pi_1$ and $Sigma_1$ formulas for $E$:
$exists k varphi_0(k,n),exists k varphi_1(k,n)$ are $Sigma_1$ formulas which define $S_0$ and $S_1$.
$S_0={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n)}$
$S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_1(k,n)}$
$S_0setminus S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land forall l ~ lnot varphi_1(l,n)}$
$E={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)}$
$exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)$ is equivalent to a $Sigma_1$ formula. Now I need a $Sigma_1$ formula for $mathbb{N} setminus E$ or a $Pi_1$ formula for $E$. Can someone give me a hint how to find it?
logic first-order-logic predicate-logic recursion computability
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
$endgroup$
I'd appreciate some help for the following exercise. $Sigma_1, Pi_1$ and $Delta_1$ are defined here: https://en.wikipedia.org/wiki/Arithmetical_hierarchy
I think I can prove 2. when I have 1. - can someone give me hint, how to start with the proof? I already know that all $Sigma_1$-sets are r.e. and all $Pi_1$-sets are co-r.e. and all $Delta_1$-sets are recursive.
Assume $S_0, S_1 subset mathbb{N}$ are $Sigma_1$-sets which satisfy $S_0 cup S_1 = mathbb{N}$. Prove that there is a $Delta_1$-set $E$ such that $S_0 setminus S_1 subset E subset S_0$ is satisfied.
Assume $Q_0, Q_1 subset mathbb{N}$ are $Pi_1$-sets which satisfy $S_0 cap S_1 = emptyset$. Prove that there is a $Delta_1$-set $E$ such that $Q_0 subset E$ and $Q_1 subset mathbb{N} setminus E$ are satisfied.
EDIT:
I'm trying to construct $Pi_1$ and $Sigma_1$ formulas for $E$:
$exists k varphi_0(k,n),exists k varphi_1(k,n)$ are $Sigma_1$ formulas which define $S_0$ and $S_1$.
$S_0={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n)}$
$S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_1(k,n)}$
$S_0setminus S_1={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land forall l ~ lnot varphi_1(l,n)}$
$E={nin mathbb{N}: mathbb{N}vDash exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)}$
$exists k ~ varphi_0(k,n) land exists l ~ lnot varphi_1(l,n)$ is equivalent to a $Sigma_1$ formula. Now I need a $Sigma_1$ formula for $mathbb{N} setminus E$ or a $Pi_1$ formula for $E$. Can someone give me a hint how to find it?
logic first-order-logic predicate-logic recursion computability
logic first-order-logic predicate-logic recursion computability
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
edited Jan 6 at 21:29
thehardyreader
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
asked Jan 5 at 22:55
thehardyreaderthehardyreader
31829
31829
add a comment |
add a comment |
1 Answer
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$begingroup$
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $ninmathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $nnotin E,$ whereas if it appears first in the $S_0$ enumeration, $nin E.$ (And break ties however.)
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
$endgroup$
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Do you have suggestions how to formalize what I wrote?
$endgroup$
– thehardyreader
Jan 6 at 11:22
$begingroup$
No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
$endgroup$
– spaceisdarkgreen
Jan 6 at 18:00
$begingroup$
Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
$endgroup$
– thehardyreader
Jan 6 at 19:04
1
$begingroup$
It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
$endgroup$
– spaceisdarkgreen
Jan 6 at 22:50
|
show 2 more comments
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$begingroup$
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $ninmathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $nnotin E,$ whereas if it appears first in the $S_0$ enumeration, $nin E.$ (And break ties however.)
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
$endgroup$
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Do you have suggestions how to formalize what I wrote?
$endgroup$
– thehardyreader
Jan 6 at 11:22
$begingroup$
No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
$endgroup$
– spaceisdarkgreen
Jan 6 at 18:00
$begingroup$
Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
$endgroup$
– thehardyreader
Jan 6 at 19:04
1
$begingroup$
It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
$endgroup$
– spaceisdarkgreen
Jan 6 at 22:50
|
show 2 more comments
$begingroup$
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $ninmathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $nnotin E,$ whereas if it appears first in the $S_0$ enumeration, $nin E.$ (And break ties however.)
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
$endgroup$
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Do you have suggestions how to formalize what I wrote?
$endgroup$
– thehardyreader
Jan 6 at 11:22
$begingroup$
No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
$endgroup$
– spaceisdarkgreen
Jan 6 at 18:00
$begingroup$
Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
$endgroup$
– thehardyreader
Jan 6 at 19:04
1
$begingroup$
It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
$endgroup$
– spaceisdarkgreen
Jan 6 at 22:50
|
show 2 more comments
$begingroup$
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $ninmathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $nnotin E,$ whereas if it appears first in the $S_0$ enumeration, $nin E.$ (And break ties however.)
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
$endgroup$
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $ninmathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $nnotin E,$ whereas if it appears first in the $S_0$ enumeration, $nin E.$ (And break ties however.)
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
answered Jan 5 at 23:45
spaceisdarkgreenspaceisdarkgreen
33.9k21754
33.9k21754
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Do you have suggestions how to formalize what I wrote?
$endgroup$
– thehardyreader
Jan 6 at 11:22
$begingroup$
No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
$endgroup$
– spaceisdarkgreen
Jan 6 at 18:00
$begingroup$
Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
$endgroup$
– thehardyreader
Jan 6 at 19:04
1
$begingroup$
It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
$endgroup$
– spaceisdarkgreen
Jan 6 at 22:50
|
show 2 more comments
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Do you have suggestions how to formalize what I wrote?
$endgroup$
– thehardyreader
Jan 6 at 11:22
$begingroup$
No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
$endgroup$
– spaceisdarkgreen
Jan 6 at 18:00
$begingroup$
Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
$endgroup$
– thehardyreader
Jan 6 at 19:04
1
$begingroup$
It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
$endgroup$
– spaceisdarkgreen
Jan 6 at 22:50
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
$endgroup$
– thehardyreader
Jan 6 at 11:07
$begingroup$
Thanks for your answer! So I can basically give a recursive characteristic function $chi_E$ for $E$: $chi_E(x)=1$ if $x in S_0$ or $x in S_0 cap S_1$ $chi_E(x)=0$ else. This function is recursive because there is an algorithm which holds in finite time if $x in S_0$or $x in S_1$. Hence $E$ is a $Delta_1$-set which also satisfies the condition $S_0 setminus S_1 subset E subset S_0$.
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– thehardyreader
Jan 6 at 11:07
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Do you have suggestions how to formalize what I wrote?
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– thehardyreader
Jan 6 at 11:22
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Do you have suggestions how to formalize what I wrote?
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– thehardyreader
Jan 6 at 11:22
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No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
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– spaceisdarkgreen
Jan 6 at 18:00
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No, $E$ probably only contains some of $S_0cap S_1$ (whichever happen to appear in the enumeration of $S_0$ before they appear in the enumeration of $S_1$). Yes, $E$ is defined in terms of some fixed recursive enumerations of the two sets, and the definition is in the form of an algorithm that decides membership in $E$ based on these, so $E$ is recursive. Not sure how formal you have to be here. If you wish, you can avoid arguments about recursiveness and use the idea I gave to write down a $Delta_1$ formula based on the $Sigma_1$ formulas defining $S_0$ and $S_1.$
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– spaceisdarkgreen
Jan 6 at 18:00
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Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
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– thehardyreader
Jan 6 at 19:04
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Okay, I see what's wrong in my comment above. What do you mean by $Delta_1$ formula? For a $Delta_1$-set there is a $Sigma_1$-formula and $Pi_1$-formula which defines it, or am I mistaken? But maybe I can do just that, writing down both for $E$.
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– thehardyreader
Jan 6 at 19:04
1
1
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It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
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– spaceisdarkgreen
Jan 6 at 22:50
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It doesn't seem like you're grasping what $E$ is. The logic above suggests that $E$ should be defined by something like $exists k (varphi_0(k,n)land forall (l< k)lnotvarphi_1(l,n))$ (i.e. it appears in $S_0$ before (or same time as) it appears in $S_1$). And the complement of $E$ can be defined something like $exists k(varphi_1(k,n)land forall (lle k) lnot varphi_0(l,n)$ (i.e. it appears in $S_1$ before it appears in $S_0$). That the former is equivalent to the negation of the latter will require the fact that $forall n(exists k varphi_0(k,n)lor exists k varphi_1(k,n))$
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– spaceisdarkgreen
Jan 6 at 22:50
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