YES

Problem:
 active(f(f(X))) -> mark(c(f(g(f(X)))))
 active(c(X)) -> mark(d(X))
 active(h(X)) -> mark(c(d(X)))
 active(f(X)) -> f(active(X))
 active(h(X)) -> h(active(X))
 f(mark(X)) -> mark(f(X))
 h(mark(X)) -> mark(h(X))
 proper(f(X)) -> f(proper(X))
 proper(c(X)) -> c(proper(X))
 proper(g(X)) -> g(proper(X))
 proper(d(X)) -> d(proper(X))
 proper(h(X)) -> h(proper(X))
 f(ok(X)) -> ok(f(X))
 c(ok(X)) -> ok(c(X))
 g(ok(X)) -> ok(g(X))
 d(ok(X)) -> ok(d(X))
 h(ok(X)) -> ok(h(X))
 top(mark(X)) -> top(proper(X))
 top(ok(X)) -> top(active(X))

Proof:
 WPO Processor:
  algebra: Sum
  weight function:
   w0 = 0
   w(ok) = 6
   w(mark) = w(active) = 4
   w(h) = 1
   w(top) = w(proper) = w(d) = w(c) = w(g) = w(f) = 0
  status function:
   st(top) = st(ok) = st(proper) = st(h) = st(d) = st(mark) = st(c) = st(
   g) = st(active) = st(f) = [0]
  precedence:
   active > proper > h ~ c ~ g ~ f > d > top ~ ok ~ mark
  problem:
   
  Qed