YES

Problem:
 h(f(x,y)) -> f(f(a(),h(h(y))),x)

Proof:
 DP Processor:
  DPs:
   h#(f(x,y)) -> h#(y)
   h#(f(x,y)) -> h#(h(y))
  TRS:
   h(f(x,y)) -> f(f(a(),h(h(y))),x)
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {5,3,1}
    transitions:
     f1(9,2) -> 10*
     f1(10,8) -> 6*
     a1() -> 9*
     h1(2) -> 2*
     h{#,1}(2) -> 3,1
     f40() -> 2*
     h{#,0}(2) -> 3,1
     h{#,0}(4) -> 3*
     h0(2) -> 4*
     h0(4) -> 6*
     f0(8,2) -> 5*
     f0(8,8) -> 6*
     f0(7,6) -> 8*
     a0() -> 7*
     3 -> 1*
     5 -> 4*
   problem:
    DPs:
     
    TRS:
     h(f(x,y)) -> f(f(a(),h(h(y))),x)
   Qed