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Of course, this depends on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on the language $A$. For instance, if $A=HP$, the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., that $M$ is a decider, $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.

Of course, this depends on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on the language $A$. For instance, if $A=HP$, the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., that $M$ is a decider, $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.

Of course, this depends on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on the language $A$. For instance, if $A=HP$, the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., that $M$ is a decider, $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.

Of course, this depends on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on the language $A$. For instance, if $A=HP$, the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., that $M$ is a decider, $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.

Of course, this is depend on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on $A$. For instance, if $A=HP$ the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.

Of course, this depends on what you exactly mean.

Do you mean, all the machines that decides a specific language? e.g., $$ L = \{ \langle M \rangle \mid M \text{ decides the language } A\}$$

then, it depends on the language $A$. For instance, if $A=HP$, the halting problem, then $L$ is clearly decidable (i.e., it is empty).

But if you mean, any language, i.e., that $M$ is a decider, $$ L = \{ \langle M \rangle \mid M \text{ halts on all inputs } \}$$ then $L$ is not recognizable, see Yuval's answer.