There are a couple of requirements given in the exercise. For given formulas S1 and S2, you have to find formulas phi1 and phi2 such that

- phi1 <-> phi2 is not valid
- phi1 <-> false is not valid
- phi2 <-> false is not valid
- (phi1 -> S1&!S2 and phi2 -> S1&!S2 are both valid) or (phi1 -> !S1&S2 and phi2 -> !S1&S2 are both valid)

Now, check these requirements for your formulas

S1 := X b & G c
S2 := G b & F c
phi1 := G (b & [c SB b])
phi2 := G (b & [c WB b])

These formulas satisfy all requirement 4, but none of 1-3 since you can prove that

(G (b & [c SB b]) ) <-> (G (b & [c WB b]))

is valid. That is, you just found one example instead of two. Syntactically they are different, but they are equivalent. Moreover, both formulas are no candidates since you can also prove that

!(G (b & [c SB b]) )

is valid, and therefore your formulas phi1 and phi2 are both false. You can see that also by unrolling the SB operator:

[c SB b] <-> !b & (c | X[c SB b])

and therefore

(b & [c SB b]) <-> b & !b & (c | X[c SB b])

which is obviously false. Therefore, none of your formulas phi1 and phi2 are candidates for a solution.