Graduate Is there a decidable set theory?

Click For Summary
The discussion centers on the undecidability of standard axiomatizations in mathematics, specifically Peano arithmetic and ZFC set theory, as highlighted by Gödel's theorem. While Presburger's axiomatization offers a decidable alternative for arithmetic, it is too weak for comprehensive arithmetic applications. The question arises whether a decidable set theory exists, with the consensus suggesting that it would likely be too weak for ordinary set theory. Additionally, the conversation touches on the implications of proof lengths and recursive functions in relation to decidability within reasoning systems. Ultimately, the exploration of decidable set theories remains open, with concerns about their limitations in expressing fundamental mathematical concepts.
Demystifier
Science Advisor
Insights Author
Messages
14,608
Reaction score
7,219
The Godel theorem shows that the standard Peano axiomatization or arithmetic is undecidable. However, there is an alternative Presburger's axiomatization of arithmetic, which is decidable.

Similarly, the standard ZFC axiomatization of set theory is undecidable. For instance, the continuum hypothesis is undecidable in ZFC. Is there an alternative axiomatization of set theory which is decidable?
 
Physics news on Phys.org
Presburger's is decidable, but it is too weak to do ordinary arithmetic. I suspect that if it exists a decidable set theory would be too weak to do ordinary set theory.
 
By Presburger arithmetic you cannot state certain general theorems, but I think you can still do arithmetic in a practical sense. (For instance, you can compute concrete products such as ##2\times 6=12##. Any 7 year old kid can do it without knowing Peano axioms.) Perhaps something similar might be true with decidable set theory.
 
Last edited:
  • Like
Likes Demystifier
While this does not answer the actual technical part of your question, it is probably of some (tangential) relevance (because of being related to decidability and proof length).

The somewhat trivial (and non-technical) observation is that consider all well-formed statements (that are suppose to have true or false values) of length n that can be formed in some "reasonable" reasoning system. Also suppose that all the well-formed statements are also decidable in the system (using some agreed and/or given rules of inference as part of the reasoning system).

For some statement S, suppose we define the minimum proof-length as the length of the shortest proof of S. Let T(n) denote the maximum value among the minimum proof-lengths for all statements of length n or less. Then T(n) has to be strictly bounded by some recursive function.

For example, think of example of Presburger arithmetic mentioned above. Then I am pretty sure there is a recursive function that would bound the minimum proof-lengths (as mentioned in previous paragraph) for a given length of statements cast in the system (as questions with truth values). And I would suspect that recursive function wouldn't really grow that much faster than much of elementary functions encountered in normal math (possibly hovering somewhere little above or below (more likely below) ackermann function).

So the inability of a "reasonable" reasoning system (PA should count I think because it can't answer some of its own questions) to cast questions that it itself can't answer can also be seen in this way ... The reasoning system should fail to prove certain statements of length n or less (that it can cast as questions) because it would completely stop proving "any" statement of length n or less after f(n) number of steps (where f is some recursive function). I don't have any good idea what that function would be for PA though.
 
SSequence said:
So the inability of a "reasonable" reasoning system (PA should count I think because it can't answer some of its own questions) to cast questions that it itself can't answer can also be seen in this way ... The reasoning system should fail to prove certain statements of length n or less (that it can cast as questions) because it would completely stop proving "any" statement of length n or less after f(n) number of steps (where f is some recursive function). I don't have any good idea what that function would be for PA though.
Sorry I think this particular part can't be assumed beforehand (under the more generic assumption of theorems being recursively enumerable).
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K