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Is there computable model of finite fragments of ZFC?

Undecidability in ZFCFinite fragments of ZFC$ V_ kappa$ ( $ kappa $ inaccessible ) models there is a countable model of ZFCIs there a countable transitive model satisfying the same set of first-order sentences as $V$?For transitive model $M$, is $L^M$ really a model of ZFC?Exercise with transitive models of ZFCMinimal model of ZFC without power set axiomConsistency strength of a proposition about models of ZFCExistence of an inner model of ZFC implies $Con(ZFC)$?Arithmetic statements independent of ZFC and standard model of arithmetic

0

$begingroup$

There isn't a computable model of $ZFC$.

Because if there is it, then it produces a non-standard computable model of $PA$, but there isn't this model because of the tennenbaum's theorem.

However, robinson arithmetic $Q$ don't satisfies the tennenbaum's theorem.(ref.)

Thus we take a finite fragments of $ZFC$ such that it is with robinson arithmetic $Q$ and it is without $PA$.

Question:
Is there a computable model of this fragments?
If there is it, then how strong can it?

Thanks in advance.

set-theory

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edited 16 hours ago

hina logic

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  So you're basically asking: how strong can a finite fragment of ZFC be while still having a computable model?
  $endgroup$
  – Noah Schweber
  15 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

There isn't a computable model of $ZFC$.

Because if there is it, then it produces a non-standard computable model of $PA$, but there isn't this model because of the tennenbaum's theorem.

However, robinson arithmetic $Q$ don't satisfies the tennenbaum's theorem.(ref.)

Thus we take a finite fragments of $ZFC$ such that it is with robinson arithmetic $Q$ and it is without $PA$.

Question:
Is there a computable model of this fragments?
If there is it, then how strong can it?

Thanks in advance.

set-theory

share|cite|improve this question

edited 16 hours ago

hina logic

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  So you're basically asking: how strong can a finite fragment of ZFC be while still having a computable model?
  $endgroup$
  – Noah Schweber
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There isn't a computable model of $ZFC$.

Because if there is it, then it produces a non-standard computable model of $PA$, but there isn't this model because of the tennenbaum's theorem.

However, robinson arithmetic $Q$ don't satisfies the tennenbaum's theorem.(ref.)

Thus we take a finite fragments of $ZFC$ such that it is with robinson arithmetic $Q$ and it is without $PA$.

Question:
Is there a computable model of this fragments?
If there is it, then how strong can it?

Thanks in advance.

set-theory

share|cite|improve this question

edited 16 hours ago

hina logic

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 16 hours ago

hina logic

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 16 hours ago

hina logic

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 17 hours ago

hina logichina logic

11

New contributor

hina logic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 17 hours ago

hina logichina logic

11