MAYBE

Problem:
 strict:
  b(p(b(x1))) -> b(q(b(x1)))
 weak:
  1(p(0(1(0(x1))))) -> p(x1)
  q(x1) -> 0(q(0(x1)))
  q(x1) -> 1(q(1(x1)))
  q(x1) -> 0(p(0(x1)))
  q(x1) -> 1(p(1(x1)))

Proof:
 Open