User blog comment:Ytosk/Trying to define Bowers' K(n) systems/@comment-35470197-20191031094256/@comment-38817512-20191101061449

Yes, the "Let x be y" means that when i say x, i mean anything that is y.

>a consistent axiomatic system without any specific structure related to arithmetic does not have a way to define a natural number.

Set theory works with sets, which clearly aren't numbers, but it can define natural numbers. That's because there is a set of things it can define, which can be well-ordered. Now imagine a completely different axiomatic system. It has a binary relation r(a,b), and the axioms are 1. There exists an x, such that for all y, r(x,y) 2. For all x, r(x,x) 3. For all x, there exists a y, such that ¬r(y,x) and for all z, (r(y,z)⇔r(x,z)) or for all w, (r(x,w)⇔r(z,w))

Here, we can well-order everything by the relation r(a,b), and thus we can think of r(a,b) as a is less than or equal to b, and the things that we can prove to exist would be the natural numbers. Now we can define a succesor function, similarly to the third axiom, but without the "there exists a y, such that" and replacing every y with s(x). We could use f(0,x) instead of s(x), and then define the fast growing hierarchy, but since we only have the natural numbers, we can only have f(x,y) for finite x. I have made an extension of this axiomatic system, which goes up to epsilon 0, but it's already quite complicated, and has over 200 symbols, while this only has 71. So, using what i've said before, let the set of natural numbers in an axiomatic system s be a well-ordered set of objects that can be proven to exist in s, and let the natural numbers be the elements of that set.

>Moreover, if you are just referring to proof-theoretic definability, this approach does not yield a large number significantly greater than Rayo's number, because it can be done by a first order oracle.

I did not understand any of that, so i guess i still have a lot to learn in googology, unless you just flexed by using unnecessarily complex words.

>you need to understand truth predicate.

Yes, i do. Thanks for reminding me.