YES

TRS:

  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))

linear polynomial interpretations on N:

  active_A(x1) = x1 + 10
  active#_A(x1) = x1 + 13
  f_A(x1) = x1 + 1
  f#_A(x1) = x1 + 3
  mark_A(x1) = x1 + 4
  mark#_A(x1) = x1 + 5
  c_A(x1) = x1 + 3
  c#_A(x1) = x1 + 5
  g_A(x1) = x1 + 3
  g#_A(x1) = x1 + 5
  d_A(x1) = x1 + 1
  d#_A(x1) = x1 + 1
  h_A(x1) = x1
  h#_A(x1) = x1 + 2
  proper_A(x1) = x1 + 1
  proper#_A(x1) = x1 + 4
  ok_A(x1) = x1 + 11
  ok#_A(x1) = 0
  top_A(x1) = 0
  top#_A(x1) = x1 + 5

precedence:

  top > active > g > f > mark > proper > h > d > ok > c