YES

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

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