Stopping calculation in nested fixed point if inner variable is not dependent on outer variable.

Question

S2=mu x. ([] x | nu y. (b & b & [] y))

Over here can we stop the calculation on the second round for the inner brackets since we can notice that it is not dependent on x and directly continue with the previous fixed point result? 

Will the answer be considered in the exam?

Answer 1

Since it is correct, it will certainly be accepted.