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Theories of arithmetic in first-order logic, such as Peano arithmetic, second-order arithmetic, Heyting arithmetic, and their subsystems and extensions. Models of Peano arithmetic.
2
votes
Accepted
To which arithmetic\set theory this theory is bi-interpretable?
Your theory is true in the one-element universe $\{a\}$ in which $a<a$ is true and $a\in a$ is false. The order transitivity holds trivially in this case; the finiteness axiom holds vacuously; and the …
24
votes
Accepted
Which recursively-defined predicates can be expressed in Presburger Arithmetic?
Presburger arithmetic admits elimination of quantifiers, if one expands the language to include truncated minus and the unary relations for divisibility-by-2, divisibility-by-3 and so on, which are de …
3
votes
The Dirichlet principle and arithmetical induction
The theory PA fully proves the pigeon-hole principle, which how I like to refer to your principle, the claim that there is no injection from the predecessors of $n$ to those of $n-1$. (This implies of …
6
votes
Accepted
Can PA define functions related to higher theories?
Yes, this function is obviously definable in PA and PA proves it is total. You are defining the tower of theories by recursion, which PA can do, and taking the Rosser sentence of each theory, which PA …
4
votes
Heuristic interpretations of the PA-unprovability of Goodstein's Theorem
To my way of thinking, the arguments you mention seem not to distinguish sufficiently between the content of Goodstein's theorem as a universal claim $\forall p$ and Goodstein's theorem as a collectio …
11
votes
Accepted
Which part(s) of this proof of Goodstein's Theorem are not expressible in Peano arithmetic?
Because your argument involves arithmetical classes at several points, as you noticed, it is not directly expressible in the first-order language of $\newcommand\PA{\text{PA}}\PA$, although as Noah me …
39
votes
Accepted
What is known about the theory of natural numbers with only 0, successor and max?
This theory is equivalent to the theory of a discrete linear order with a least element and no largest element, that is, the theory of $\langle\mathbb{N},<\rangle$. From max we can define the order an …
10
votes
Accepted
Why include $0$ and $1$ in the signature of Presburger arithmetic?
It is the same in Peano arithmetic, where the standard language is $\{+,\cdot,0,1,<\}$ for the standard model $\langle\mathbb{N},+,\cdot,0,1,<\rangle$, even though $0$, $1$, and $<$ are definable from …
4
votes
Is this theory synonymous with PA?
Let me give half an answer.
Specifically, we can interpret your theory in PA via the Ackermann encoding. Namely, in any model of PA, define $m\in x$ if and only if the $m$th bit of $x$ is $1$. This fu …
2
votes
Is diamond consistent with 2nd order PA?
In the case of $\text{GBC}^-$, the answer is no, $\Diamond$ is not provable from the assumption that all reals are constructible. It is consistent with $\text{GBC}^-$ and even $\text{KM}^-$ that $\Dia …
16
votes
Con(PA) via non-well-foundedness?
This is a completely standard perspective in work on models-of-PA, a view that informs dozens of arguments. That is simply the nature of nonstandard models, that things they think are well founded are …
14
votes
Accepted
Extensions of $PA+\neg Con(PA)$ with large consistency strength
Let $T$ be the theory PA + ¬Con(PA), plus the axiom asserting that there is no proof of a contradiction in ZFC (or ZFC+LC etc.) of size below $k$, where $k$ is the smallest such that $\newcommand\PA{\ …
8
votes
Can set theory be interpreted in infinite arithmetic?
Without considering your system of arithmetic too closely, let me mention that ZFC is interpretable in Peano arithmetic, if one augments PA with the assertion that ZFC is consistent.
That is, ZFC is i …
7
votes
Accepted
The additive structure of clusters of nonstandard models of arithmetic
The answer is yes. The additive cluster structure knows the additive structure of the original model up to isomorphism.
Theorem. The additive structure of any countable nonstandard model $M$ of $\newc …
7
votes
The additive structure of clusters of nonstandard models of arithmetic
If you intend literally to recover the addition operation of the given model $M$, then the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly …