MAYBE

Problem:
 strict:
  a(x1) -> x1
  b(x1) -> x1
  a(c(x1)) -> b(b(c(a(x1))))
  d(b(b(b(x1)))) -> a(d(a(x1)))
 weak:
  a(b(x1)) -> b(a(x1))
  b(a(x1)) -> a(b(x1))

Proof:
 Open