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Let's take a relatively strong axiomatic system, call it $A$. By Godel's Incompleteness Theorem, there are infinitely many true statements expressible in $A$ that cannot be proven. Let's call the set of all truths in $A$ that are provable, $T$. And then all nonprovable truths $NT$.

If we have another axiom system where all the axioms are the negations of $A$'s axioms, i.e $\neg A$, the set of provable theorems is trivially $\neg T$. My question is if the same applies to $NT$, if the set of all unprovable statements in $\neg A$ is $\neg NT$.

In simpler terms, are all unprovable statements in the negation of an axiomatic system the negation of an unprovable statement in the original system?

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    $\begingroup$ Are you sure that "the set of provable theorems is trivially $\neg T$"? To take an extreme example, what if one of the axioms in $A$ is a tautology? $\endgroup$ – Chris Culter Jul 13 '17 at 18:31
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Your second paragraph is incorrect. We can have a consistent theory $A$ such that $\neg A$ is inconsistent: this means that $\neg A$ proves everything, not just the negations of the theorems of $A$. Similarly, if $A$ is inconsistent but $\neg A$ is consistent, then everything is a negation of a theorem of $A$, so $\neg A$ does not prove all negations of theorems of $A$. There are plenty of further examples. For instance, take $p\in A$ and $\neg q\in A$. Then the sentence "$p\vee q$" is a theorem of both $A$ and $\neg A$, even if it isn't a tautology.

So in general there is no obvious relationship between the theorems of $A$ and the theorems of $\neg A$. Similarly, the answer to the question you ask is "no."

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    $\begingroup$ That is unintuitive but it makes sense. Thank you very much! $\endgroup$ – TreFox Jul 13 '17 at 18:43

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