The Second Incompleteness Theorem says that if $T$ is a consistent (computably) axiomatizable theory which extends IΣ1, then $\mathrm{Con}(T)$ is not provable from $T$. By analogy with computability theory, the stronger theory $T + \mathrm{Con}(T)$ can be thought of as the "jump" of $T$. To abuse this analogy, I will use $T'$ to denote the theory $T + \mathrm{Con}(T)$. I will write $T \leq S$ when $S$ proves every axiom of $T$; I will also write $S \equiv T$ (resp. $T < S$) when $T \leq S$ and $S \leq T$ (resp. $S \nleq T$). 


It is well-known that if $T$ is consistent there are plenty of axiomatizable theories $S$ such that $T < S < T'$. In the following questions $H$ will denote an operator (like $\mathrm{Con}$) that uses the computable axiomatization of $T$ to produce a sentence $H(T)$. I will write $T^H$ for the theory $T + H(T)$.


1. Is there a computable operator $H(T)$ such that $T < T^H < T'$ for every consistent axiomatizable theory $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?
2. Is there a computable operator $H(T)$ such that $(T^H)^H \equiv T'$ for every consistent axiomatizable $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?


Question 1 asks for a uniform solution to the analogue of Post's Problem for axiomatizable theories. Question 2 asks for a uniform "half-jump" operator.



Uniform solutions to Post's problem for axiomatizable theories
(Note: this has been rewritten to reflect the comments below). 


The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive. 


Start with a (consistent) theory T to which the second incompleteness theorem applies, which means that T + ~Con(T) is also consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$~S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$S) is consistent, so T + Con(T)$\lor$S is stonger than T. 


If T $\vdash$ (Con(T)$\lor$S) $\to$ Con(T) then T $\vdash$ S $\to$ Con(T). But this means T $\vdash$ ~Con(T) $\to$ ~S which is impossible. This shows that T + (Con(T)$\lor$S) < T+ Con(T) .


So we can let TH be T + (Con(T)$\lor$S). 


