Definitely not. Look, the sub formula behind the G-operator is false, since it states that all all points of time, except for the initial one, we should have (!a&!b&c) <-> (!a&!b&!c), but c cannot be true and false at the same time.
Let us first define
- s0 := !a&!b&!c
- s1 := !a&!b&c
- s2 := !a&!b&!c ( = s0)
Then, I would write instead something like
s0 & (X s1) & (X X s2) & X G ((s1 -> X s2) & (s2 -> X s1))
The above can be generalized to any kind of counterexample that you will be given by the LTL prover. It describes an initial prefix that is followed by a cycle.
In the above case, this is however not required, since s0=s2 holds (the LTL tool does not generated the smallest possible counterexample). Hence, it should be sufficient to write
s0 & G((s0 -> X s1) & (s1 -> X s0))
