Can we have a theory $T$ that is complete for simple sentences in the language of $T$ that are weaker than $ Con(T)$?

Question

Let's denote a sentence $P$ as "weak Godel sentence of theory $T$", if and only if $$[\neg (T \vdash P) \wedge \neg (T\vdash \neg P)] \wedge [Con(T)=Con(T+P) \wedge Con(T)=Con(T+ \neg P)] $$

In English this is: $P$ is independent of $T$ and the addtion of $P$ or $\neg P$ to $T$ doesn't prove the consistency of $T$.

Let's denote a sentence as complex if it has a proper subformula of it that is a sentence, or when de-prenexed it results in a sentence that has a proper subformula of it that is a sentence. A sentence is simple if and only if it is not complex.

Let's fix the language of $T$ to a classical first order logic language that doesn't contain any constants in its signature. By sentence it is meant the usual meaning of a fully quantified formula (i.e. has no free variables).

Definition: $$T \text{ is complete for simple sentences below } Con(T) \iff \forall P (P \text { is a weak Godel sentence of }T \to P \text { is complex})$$

In other words: all sentences if the addition of them or their negation to $T$ doesn't result in a theory that can prove the consistency of $T$, that are simple, then those sentences are decidable by $T$.

Question: is it possible to have a theory that meets Godel's incompleteness criteria and yet is complete for simple sentences below its consistency level?

Answer 1

Unless I'm missing something, every sentence is equivalent to a simple sentence. To see this, take a sentence $\varphi$ and produce an equivalent sentence $\psi$ by the following procedure: Let $x$ be a variable not occuring in $\varphi$. Replace each atomic sub-formula $\chi$ of $\varphi$ by $\chi \wedge x = x$ if $\chi$ is in the scope of an even number of negations and $\chi \vee x \neq x$ if $\chi$ is in the scope of an odd number of negations (we're treating $\alpha \rightarrow \beta$ as $\neg\alpha \vee \beta$). Finally add $\exists x$ to the beginning of the sentence. To see that they're equivalent, note that if $\psi$ holds, then we can pick a witness $c$ for the $\exists x$ quantifier and substitute it in for each instance of $x$, giving a sentence that is clearly equivalent to $\varphi$. Conversely if $\varphi$ holds, then if we let $x$ be any $c$ whatsoever, substituting in $c$ for $x$ in $\psi$ with the first quantifier removed again gives a sentence clearly equivalent to $\varphi$. Finally, every proper subformula of $\psi$ fails to be a sentence, so it is simple.

Edit: Assuming that the language in question contains addition and the theory proves basic facts about arithmetic, there's a more robust way of converting any sentence $\varphi$ into a simple sentence, namely let $x$ and $y$ be variables not appearing in $\varphi$ and let $\varphi^\prime$ be $\varphi$ with each term $t$ replaced with $t+x$. Then an equivalent simple sentence is $\exists x ((\forall y (y = y + x)) \wedge \varphi^\prime)$.

Answer 2

Partial answer:

If there is such a theory $T$, it would mean that every simple sentence that's independent of $T$, must decide Con($T$).

Note that every $\Pi^0_1$ sentence is equivalent to a simple sentence by Matiyasevich's theorem (every $\Sigma^0_1$ set is diophantine). And I guess we assume $T$ proves Matiyasevich's theorem.

So in particular,

every true $\Pi^0_1$ sentence that's independent of $T$ must decide Con($T$).

(The false $\Pi^0_1$ sentences are refutable in $T$; I think this, with the fact that $T$ is $\Delta^0_1$-axiomatizable, is part of your assumption "meets Gödel's incompleteness criteria".)

I don't know enough about which $\Pi^0_1$ sentences decide Con($T$), but it seems like a tall order that such a theory should exist.