Does showing a problem and its complement are not Turing-decidable means that the language & its complement are not Turing-recognizable? I was reading the Sipser's book on the Theory of Computation, 3rd edition and came up with a question. "Does showing a problem and its complement are not Turing-decidable means that the language &amp; its complement are not Turing-recognizable?" I believe that the answer is NO, however, the Theorem 5.30 states something different. 

There are two problems concerned in this question. One is $A_{TM} = \{ &lt;M,w&gt;\ |\ M \text{ is a TM and accepts } w\}$ and the other is $EQ_{TM} = \{ &lt;M_1,M_2&gt;\ |\ M_1, M_2 \text{ are a TMs and } L(M_1) = L(M_2)\}$.

On Page 238, the Theorem 5.30 is stated as follows:


  Theorem 5.30 $EQ_{TM}$ is neither Turing-recognizable nor co-Turing-recognizable.


The proof is by mapping reduction of $A_{TM}$ to $\overline{EQ_{TM}}$, and at the same time, reduction from $A_{TM}$ to $EQ_{TM}$. This way, it has shown:


$\overline{EQ_{TM}}$ is Turing-undecidable.
$EQ_{TM}$ is Turing-undecidable. 


Note that this reduction does not show that either of $\overline{EQ_{TM}}$ or $EQ_{TM}$ are not Turing-recognizable, since $A_{TM}$ is Turing-recognizable.

On the other hand, on Page 209, we have the following definition


  A language is co-Turing-recognizable if it is the complement of a Turing-recognizable language.


which is accompanied by Theorem 4.22:


  Theorem 4.22 A language is decidable iff it is Turing-recognizable and co-Turing Recognizable.


Therefore, we can obtain the following Corollary: If a language is not decidable, then either the language itself, its complement or both of them are not Turing-recognizable.

Therefore, it is shown on Theorem 5.30 that $\overline{EQ_{TM}}$ and $EQ_{TM}$ are Turing-undecidable. This does not tell me anything more than the following proposition:


  Either $EQ_{TM}$ or $\overline{EQ_{TM}}$ or both are not Turing-recognizable.


However, the Theorem 5.30 states something stronger. 

Am I missing something?
... a problem and its complement are not Turing-decidable means that the language &amp; its complement are not Turing-recognizable?  


This claim is false. In fact, at least one of them (or both are) is Turing-recognizable. Assume that both the set (problem) $A$ and its complement $\overline{A}$ are not decidable. Then if both $A$ and $\overline{A}$ were recognizable (recursively enumerable) then that would mean $A$ and hence $\overline{A}$ are decidable (Theorem 4.22), a contradiction.   


  ... however, the Theorem 5.30 states something different.


This theorem states that $EQ_{TM}$ is neither Turing-recognizable nor co-Turing-recognizable. In other words, both $EQ_{TM}$ and $\overline{EQ}_{TM}$ are not Turing-recognizable (recursively enumerable). This does not contradict to the above-stated claim.

Also, pay attention that the theorem 5.30 reduces $A_{TM}$ to $\overline{EQ}_{TM}$ (not to $EQ_{TM}$ as you claim in your post)  meaning that $EQ_{TM}$ cannot be Turing-recognizable (otherwise $\overline{A}_{TM}$ would be Turing-recognizable and hence decidable).      

You are right, if a language is not decidable, then either the language itself, its complement or both of them are not Turing-recognizable, and since the theorem 5.30 proves that $EQ_{TM}$ is not decidable, it follows that either $EQ_{TM}$ either, $\overline{EQ}_{TM}$ or both of them are not Turing-recognizable. But this result is weaker and does not contradict to the theorem 5.30. In fact the latter (weaker) result is not necessary since the theorem already establishes the stronger result.