Axiomatized Formal Theory Proof

  • Context: MHB 
  • Thread starter Thread starter agapito
  • Start date Start date
  • Tags Tags
    Proof Theory
Click For Summary
SUMMARY

Axiomatized Formal Theory (AFT) is proven to be undecidable when it is consistent and sufficiently strong, as established through the application of the Diagonal Argument. Key theorems supporting this conclusion include Church's theorem, which states that pure first-order predicate calculus is undecidable, and the undecidability of Robinson's system $Q$. These results can be found in resources such as the Open Logic Project's "Sets, Logic, Computation" and "Incompleteness and Computability" by Boolos et al.

PREREQUISITES
  • Understanding of Axiomatized Formal Theory (AFT)
  • Familiarity with Church's theorem and its implications
  • Knowledge of Turing machines and their encoding in predicate calculus
  • Basic concepts of Gödel's incompleteness theorems
NEXT STEPS
  • Study the Diagonal Argument and its application in formal proofs
  • Explore Church's theorem in detail and its consequences for decidability
  • Investigate Robinson's system $Q$ and its undecidability proofs
  • Read "Incompleteness and Computability" by Boolos et al. for comprehensive insights
USEFUL FOR

Mathematicians, logicians, computer scientists, and anyone interested in the foundations of formal theories and their decidability properties.

agapito
Messages
46
Reaction score
0
Can someone please direct me to ,or show, a proof that a Consistent and Sufficiently Strong AFT is not decidable. It presumably involves the Diagonal Argument, but I can't figure out how to apply it.

Many thanks.
 
Technology news on Phys.org
What is AFT?

Also, "formal" in the context of proof lately means proofs that have been verified by a computer program (proof assistant, such as Coq, Isabelle, Agda and so on). Do you have a formalized proof in mind or simply a proof from a textbook?
 
I realized that AFT stands for Axiomatized Formal Theory.

There is Church's theorem that says that pure first-order predicate calculus is not decidable. This is proved by showing that Turing machines can be encoded in predicate calculus and the fact that a machine halts can be expresses as a formula. There is also a theorem that Robinson's system $Q$ (Peano Arithmetic without induction) is undecidable. This is proved by showing that computable functions are representable in $Q$. Again, for each machine and input there exists a formula saying that the machine halts on that input, and the machine indeed halts iff that formula is provable.

These two results can be found, for example, in the Open Logic Project books "Sets, Logic, Computation" and "Incompleteness and Computability". Another comprehensive book that deals with this is Computability and Logic by Boolos et al.

The diagonal argument is used, in particular, in the proof of second Gödel's theorem about incompleteness of Peano Arithmetic.
 
Thank you very much.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
889
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
620
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 50 ·
2
Replies
50
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K