Is this Approach Correct for CTL and LTL?

Question

E(Fa ∧ Fb)

Ans- EFa & EFb        As E(a & b) = Ea & Eb

 Explanation - As EFa and EFb both are CTL ,

 thus its Pure CTL.

Answer 1

No, this is not correct. The E-quantifier can (in general) not be distributed over a conjunction. Look, E(Fa ∧ Fb) means that there must be a path where at some point of time "a" holds and at some other or the same point of time "b" holds. But it must be one and the same path. 

In contrast, (EFa) ∧ (EFb) means that there is a path p1 where at some point "a" holds, and there may be another path p2 where at some point of time "b" holds. You can quickly find a Kripke structure with two paths satisfying this, but not E(Fa ∧ Fb).

But you can convert E(Fa ∧ Fb) to CTL by enumerating the orderings of "a" and "b", i.e., 

EF (a ∧ EFb) ⋁ EF (b ∧ EFa)

which is an equivalent CTL formula.