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 $T$. 



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> 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?
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Can we have a theory $T$ that is complete for simple sentences in the language of $T$ that are weaker than $ Con(T)$?
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 $\exists x ((\forall y (y = y + x)) \wedge \varphi^\prime)$.


