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first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
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An arithmetic hierarchy without bounded quantifiers
I posted this question on math.stackexchange.com with no answers (https://math.stackexchange.com/questions/4337852/an-arithmetic-hierarchy-without-bounded-quantifiers).
Let $\exists_n$ formulas in the …
2
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1
answer
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Special classes of the arithmetical hierarchy of sentences of finite-order arithmetic
We work in a countable language of finite-order arithmetic, which allows us to quantify over natural numbers, sets of natural numbers, sets of sets of natural numbers, and so on. We measure the comple …
9
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1
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What can $I\Delta_0$ prove?
What combinatorial and number-theoretic propositions can $I\Delta_0$ prove? Obviously there are an infinitude of them, but what are some well known theorems that can be proved in $I\Delta_0$, if any?
17
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Did Edward Nelson accept the incompleteness theorems?
Edward Nelson advocated weak versions of arithmetic (called predicative arithmetic) that couldn't prove the totality of exponentiation. Since his theory extends Robinson arithmetic, the incompleteness …
4
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0
answers
545
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Can Robinson arithmetic prove any interesting theorems?
The motivation for my question is I'm curious whether studying Robinson arithmetic can be fruitful in the same sense as studying group theory. Robinson arithmetic is so weak that there are many struct …
3
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Set theories that are complete modulo finite-order arithmetic
In a previous question, I asked whether there can be effectively axiomatizable set theories (at least as strong as, say, ZF) that are complete modulo first-order arithmetic, to which the answer is no; …
7
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2
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Incompleteness theorems for theories with omega-rule
Recall that the omega-rule is an infinitary rule of inference that allows one to deduce $\forall x A(x)$ from $A(0), A(1), \dots$. It's known that adjoining PA (or even Q) with the omega-rule results …