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Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that is a union of a set and its singleton.

Questions:

 
  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?

    is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?

  2. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?

  3. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that is a union of a set and its singleton.

Questions:

 
  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that is a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?

  2. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?

  3. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

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Zuhair Al-Johar
  • 10.2k
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  • 12
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Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that is a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that is a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?
added 413 characters in body
Source Link
Zuhair Al-Johar
  • 10.2k
  • 1
  • 12
  • 44

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal}$$\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when containing an element$y$ is non empty then it$y$ is a successor and every non-empty element of it$y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal}$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when containing an element then it is a successor and every non-empty element of it is a successor"

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?

Let $x^\phi$ be the set $x$ that is definable after the parameter free formula $\phi$, i.e. formally we have: $$\forall y (y \in x^{\phi} \leftrightarrow \phi(y))$$

Now by $\text{definable ZFC}$ it is meant the theory having axioms of $\text{ZFC}$ from definable parameters, so every axiom of ZFC becomes a schema, for example axiom of pairing becomes:

Definable Pairing schema: if $\phi, \psi$ are parameter free formulas, then: $$\forall a^\phi, b^\psi \exists x \ \forall y \ (y \in x \leftrightarrow y=a \lor y=b)$$, is an axiom.

Definable Separation schema would be written as: if $\phi_1,...,\phi_n$ are parameter free formulas, and if $ \psi(y,x_1 ^{\phi_1},..,x_n^{\phi_n}) $ is a formula in which only symbols $``y,x_1^{\phi_1},..,x_n^{\phi_n}" $ can occur free, and only occurring free, then:

$$\forall x_1^{\phi_1},..,x_n^{\phi_n} \exists x \forall y \ [y \in x \leftrightarrow y \in x_n^{\phi_n} \wedge \psi(y,x_1^{\phi_1},..,x_n^{\phi_n})]$$, is an axiom.

The same to be applied over all other axioms of $\text{ZFC}$, and infinity is axiomatized in a definable manner like there exists a set of all finite von Neumann ordinals. The only two axioms that are not presented in a definable manner are Extensionality and Foundation.

The formulation of axiom of infinity is the following:

Definable infinity: $\exists x \forall y (y \in x \leftrightarrow y \text{ is a finite von Neumann ordinal})$, where "$y$ is a finite von Neumann ordinal" is definable after the parameter free formula "$y$ is a transitive set of transitive sets that when $y$ is non empty then $y$ is a successor and every non-empty element of $y$ is a successor", where "transitive" means a set whose elements are subsets of it, and "a successor" means a set that a union of a set and its singleton.

Questions:

  1. is $\text{definable ZFC}$ equi-interpretable with $\text{ZFC}$?
  1. if Yes to 1, then is $\text{definable ZFC}$ bi-interpretable with $\text{ZFC}$?
  1. if No to 1, then what would be the consistency strength of $\text{definable ZFC}$?
added 413 characters in body
Source Link
Zuhair Al-Johar
  • 10.2k
  • 1
  • 12
  • 44
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Source Link
Zuhair Al-Johar
  • 10.2k
  • 1
  • 12
  • 44
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