You can apply Rice's Theorem . How ?
$L = \left \{ \left \langle M \right \rangle | TM\ halts\ on\ every\ input\ \right \} \Rightarrow L = \left \{\left \langle M \right \rangle | L(M)\ is\ Recursive \ \right \}$
Here, Implied Reduction exists. You pick a TM you know that halts on every Input and TM which halts on every input must have a Recursive (Decidable) Language.
So, this Language can be implied reduced to
$L = \left \{\left \langle M \right \rangle | L(M)\ is\ Recursive \ \right \}$
- Now, Simply apply Rice theorem 1 to prove it as Non Trivial-Property and hence, Undecidable.
- Secondly, We can even apply Rice non monotonic Property (Rice second thorem) and see that Property of Language being Recursive is a Strict Subset of RE language, making this language not even Semi - Decidable.
For Rice theorem http://gatecse.in/rices-theorem/
Exercises :-
1). $L = \left \{ \left \langle M \right \rangle | TM\ halts\ on\ No\ input\ \right \} \Rightarrow L = \left \{\left \langle M \right \rangle | L(M)\ is\ PHI \ \right \}$ ?
(Again UD and Non RE ?)
2). $L = \left \{\left \langle M \right \rangle | L(M)\ is\ Regular \ \right \}$ ?
3). $L = \left \{\left \langle M \right \rangle | L(M)\ is\ DCFL \ \right \}$ ?
4). $L = \left \{\left \langle M \right \rangle | L(M)\ is\ CFL\ \right \}$ ?
5). $L = \left \{\left \langle M \right \rangle | L(M)\ is\ CSL \ \right \}$ ?