The claim is that (Gd & (FGd | Fb)) is equivalent to Gd. To see this, note first that Gd implies FGd since Gd demands that always d holds, and hence, it then holds also after any other point of time forever. Now, abbreviate p:=Gd, q:=FGd and r:=Fb and note finally that the following propositional logic formula is valid:

(p->q) -> (p & (q | r) <-> p)