MAYBE

Problem:
 a(a(a(b(b(x1))))) -> b(b(b(x1)))
 b(a(a(a(b(x1))))) -> a(a(a(b(a(a(a(x1)))))))
 a(a(a(x1))) -> a(a(x1))
 b(b(x1)) -> a(b(a(x1)))

Proof:
 Open