Help me Prove that such a set does NOT exist

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SUMMARY

The discussion centers on the proof that no decidable set R exists such that the Godel numbers of Robinson Arithmetic Q are included in R, and the Godel numbers of the negation of sentences in Q are also included in R. The theory T is defined as "nice" if it is consistent, recursively adequate, and extends Q. The fixed-point lemma is referenced, indicating that for any nice theory T and formula φ, there exists a sentence σ such that T proves σ is equivalent to φ("σ"). The user seeks assistance in proving this theorem, particularly focusing on the relationship between R and its complement.

PREREQUISITES
  • Understanding of Robinson Arithmetic (Q)
  • Familiarity with Godel numbering and its implications
  • Knowledge of decidability in formal theories
  • Concept of fixed-point lemma in mathematical logic
NEXT STEPS
  • Study the implications of the fixed-point lemma in formal theories
  • Research Godel's incompleteness theorems and their relation to decidability
  • Explore the concept of recursive adequacy in logical theories
  • Examine proofs involving the relationship between a set and its complement in formal logic
USEFUL FOR

This discussion is beneficial for mathematicians, logicians, and computer scientists interested in formal theories, decidability, and the foundations of arithmetic. It is particularly relevant for those studying mathematical logic and Godel's theorems.

norman95
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  1. Let Q denote the theory of Robinson Arithmetic. A theory T is nice iff T is consistent, is p.r. adequate and extends Q. The fixed-point lemma states that for all nice theories T, for any formula φ, there is a sentence σ such that
    T ⊢σ↔φ("σ") Given a set of formulas X, let gn(X) = {"φ" | φ ∈ X} be the set of Godel numbers of formulas in X. (Quotation mark denote Godel numbers.) Prove that there is no decidable set R such that gn(Q) ⊆ R and gn({σ | ¬σ ∈ Q}) ⊆ R.

Please help guys, I've been really sick for the last two week and missed all the relevant material, and now this is due tonight!
 
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This is what I have so far, but I have no idea if it's anywhere close to correct and I have like 4 chapters of my textbook that I need to cover the relevant materiel. My intuition is that either I have to show some sort of overlap between R and R-completment, or that I have to show that the decideability of R entails the decideability of Q (which is a contradiction).
 

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