YES

Problem:
 active(f(X)) -> mark(g(h(f(X))))
 mark(f(X)) -> active(f(mark(X)))
 mark(g(X)) -> active(g(X))
 mark(h(X)) -> active(h(mark(X)))
 f(mark(X)) -> f(X)
 f(active(X)) -> f(X)
 g(mark(X)) -> g(X)
 g(active(X)) -> g(X)
 h(mark(X)) -> h(X)
 h(active(X)) -> h(X)

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {13,10,3,11,9,6,1}
   transitions:
    f50() -> 2*
    mark0(5) -> 1*
    mark0(2) -> 7*
    g0(2) -> 10*
    g0(4) -> 5*
    h0(7) -> 12*
    h0(2) -> 13*
    h0(3) -> 4*
    f0(7) -> 8*
    f0(2) -> 3*
    active0(10) -> 9*
    active0(12) -> 11*
    active0(8) -> 6*
    h1(70) -> 71*
    h1(40) -> 41*
    h1(20) -> 21*
    h1(62) -> 63*
    h1(64) -> 65*
    h1(34) -> 35*
    h1(86) -> 87*
    h1(76) -> 77*
    f1(80) -> 81*
    f1(32) -> 33*
    f1(74) -> 75*
    f1(19) -> 20*
    f1(56) -> 57*
    active1(25) -> 26*
    g1(24) -> 25*
    g1(21) -> 22*
    mark1(22) -> 23*
    active2(43) -> 44*
    g2(42) -> 43*
    f2(54) -> 55*
    f2(48) -> 49*
    2 -> 34,32
    4 -> 24*
    6 -> 7,19
    7 -> 19*
    8 -> 70,56
    9 -> 7,19
    10 -> 76,74
    11 -> 7,19
    12 -> 86,80
    21 -> 42*
    22 -> 62,48
    23 -> 6*
    26 -> 1*
    33 -> 20,40,8
    35 -> 12*
    41 -> 21*
    43 -> 64,54
    44 -> 23,6
    49 -> 8,20
    55 -> 8,20
    57 -> 8,20
    63 -> 12*
    65 -> 12*
    71 -> 12*
    75 -> 8,20
    77 -> 12*
    81 -> 8,20
    87 -> 12,80
  problem:
   
  Qed