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? 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?