Let$\:$ $T=\{\varphi \in \Pi\_1: PA+Con(PA) \vdash \varphi\:\:and\:\: PA\nvdash \varphi \}$. $\:$By the facts presented here [Are undecidable consequences of Con recursively enumerable?](https://mathoverflow.net/questions/63690/are-undecidable-consequences-of-con-recursively-enumerable/63697#63697) by Andreas Blass and Emil Jerabek, we know that:


$(1)$ $T$ is $\Pi\_1$-hard. 


However it seems we can also prove that 


$(2)$ $Cn(PA+Con(PA)) = Cn(PA+T)$


Let $\varphi$ be such that $PA+Con(PA) \models \varphi$. Obviously $Con(PA) \in T$, therefore trivially $PA+T \models \varphi$. 


In the other direction, let $\varphi$ be such that $PA+T \models \varphi$. Since this is a first-order theory, by completeness and compactness we can infer that in the proof of $\varphi$ from $PA+T$ we use finitely many formulae, namely: $\phi\_1, \phi\_2, \dots \phi\_n$ .
All of them either belong to $PA$ or belong to $T$ or can be inferred from $PA+T$. In particular they are implied by $PA+Con(PA)$. If so, they can be used in the proof of $\varphi$ form $PA+Con(PA)$, so $PA+Con(PA) \models \varphi$.


But from the work of Jeroslow <http://www.jstor.org/discover/10.2307/30226121?uid=3738840&uid=2&uid=4&sid=21102368601827> (Corollary 4) we find out that:


$(3)$ $Cn(PA+\{\varphi \in \Pi\_1: \mathbb{N} \models \varphi \})$ is not $\Delta\_2$


Since $(1)$ and due to the fact that the set above (I mean: $\Pi\_1 \cap Th(\mathbb{N})$) has got its own truth definition and is $\Pi\_1$ we infer that it is reducible to $T$. 


So it seems that $Cn(PA+T)$ should also be "hard" and "not learnable" (in the sense of not being $\Delta\_2$ ). 


However we also know that $Cn(PA+Con(PA))$ is $\Sigma\_1$. Since $(2)$ however $Cn(PA+T)$ is the same set of formulae. 


So my question is: did I make any mistake in the reasoning above and one of $(1)$, $(2)$, $(3)$ is false or it can be the case that when we close some "hard" set up logical consequence, we can get an "easier" set. If so, the question is: how come? 



Question about undecidable consequences of Con, learnability  and arithmetical complexity of logical consequence 
The $\Pi\_1$-hardness of $T$ comes from the clause $PA\not\vdash\varphi$. But when you consider $Cn(PA+T)$ its influence vanishes.


As 
$$T\subset Cn(PA+Con(PA))$$ we have $$Cn(PA+T)\subseteq Cn(PA+Cn(PA+Con(PA)))=Cn(PA+Con(PA)).$$ On the other hand $Con(PA)\in T$ therefore $Cn(PA+Con(PA))\subseteq Cn(PA+T)$.


Thus we have the equality $Cn(PA+T) = Cn(PA+Con(PA))$ indeed.


This also does not contradict Jeroslow's results as in $PA+\Pi\_1(\mathbb{N})$ there are all $Con(X)$ (not just $Con(PA)$) and which is a sourse of complexity of this theory.


It is not very suprising that Cn operator can decrease the complexity of a set of sentences - you can always add a negation of a sentence of any given set to obtain an inconsistent theory i.e. primitive recursive. But it is a nice example how it can decrease it to something higher than just PR.


