Can we recognize wheter a Turing machine is a decider? [duplicate]

Let

$L=\{ \langle M \rangle \mid M \text{ is a Turing Machine which halts on all inputs}\}$.

Is this a Turing-recognizable language? I guess that it is neither Turing-recognizable, nor co-Turing-recognizable, but I can't prove it.

marked as duplicate by Ran G., Luke Mathieson, David Richerby, Kaveh, JuhoJun 6 '15 at 20:34

• You can apply Rice's theorem. But more is true: in fact, $L$ is $\Pi_2$-complete. – Yuval Filmus May 15 '14 at 14:39