Translate the following formula to an equivalent µ-calculus formula.

EFEGb

According to the cheatsheet provided, I get the following answer:

EF(**ν**x.*b* ∧ ♦*x*)

**µ**x.(**ν**y.♦*y *∧ (**ν**x.(b ∧ ♦*x*)) ∨ ♦*x*)

The solution provided is:

**µ**y.(**ν**x.♦*x** *∧ (**ν***x.*(*b *∧ ♦*x*)) ∨ ♦*y)*

Are both of these equal and correct?