YES

Problem:
 a(b(x1)) -> d(x1)
 b(a(x1)) -> a(b(x1))
 d(c(x1)) -> f(a(b(b(c(x1)))))
 d(f(x1)) -> f(a(b(x1)))
 a(f(x1)) -> a(x1)

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {11,10,5,3,1}
   transitions:
    f50() -> 2*
    a0(8) -> 9*
    a0(4) -> 3*
    a0(2) -> 11*
    d0(2) -> 1*
    a1(22) -> 23*
    a1(24) -> 25*
    d1(12) -> 13*
    d1(18) -> 19*
    b0(2) -> 4*
    b0(6) -> 7*
    b0(7) -> 8*
    c0(2) -> 6*
    f0(9) -> 5*
    f0(3) -> 10*
    19 -> 9*
    7 -> 18*
    2 -> 12*
    13 -> 3*
    1 -> 11*
    5 -> 13,1,3
    9 -> 22*
    3 -> 24,4
    10 -> 13,1,3
    25 -> 3,4
    23 -> 25,3,4
  problem:
   
  Qed