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: 2
   enrichment: match-dp
   automaton:
    final states: {12}
    transitions:
     h{#,1}(10) -> 9,11,12
     h{#,1}(7) -> 9,11,12
     h{#,1}(14) -> 12,9,11
     h{#,1}(9) -> 9,11,12
     h{#,1}(8) -> 9,11,12
     h1(10) -> 14*
     h1(7) -> 14*
     h1(14) -> 15*
     h1(9) -> 14*
     h1(8) -> 14*
     f1(17,7) -> 18,14
     f1(17,9) -> 18,14
     f1(17,17) -> 15*
     f1(16,15) -> 17*
     f1(17,8) -> 18,14
     f1(17,10) -> 18,14
     a1() -> 16*
     h{#,2}(10) -> 9,11,12
     h{#,2}(7) -> 9,11,12
     h{#,2}(9) -> 9,11,12
     h{#,2}(18) -> 9,11,12
     h{#,2}(8) -> 9,11,12
     h{#,0}(10) -> 12,9
     h{#,0}(7) -> 12,9
     h{#,0}(9) -> 12,9
     h{#,0}(13) -> 12*
     h{#,0}(8) -> 12,9
     h2(10) -> 18*
     h2(7) -> 18*
     h2(9) -> 18*
     h2(8) -> 18*
     f0(8,7) -> 7,13,8
     f0(8,9) -> 7,13,8
     f0(9,8) -> 8*
     f0(9,10) -> 8*
     f0(10,7) -> 8*
     f0(10,9) -> 8*
     f0(7,7) -> 8*
     f0(7,9) -> 8*
     f0(8,8) -> 7,13,8
     f0(8,10) -> 7,13,8
     f0(9,7) -> 8*
     f0(9,9) -> 8*
     f0(10,8) -> 8*
     f0(10,10) -> 8*
     f0(7,8) -> 8*
     f0(7,10) -> 8*
     h0(10) -> 7*
     h0(7) -> 7*
     h0(9) -> 7*
     h0(11) -> 13*
     h0(8) -> 7*
     a0() -> 10*
     7 -> 11*
     8 -> 11*
     9 -> 11*
     10 -> 11*
   problem:
    DPs:
     h#(f(x,y)) -> h#(y)
    TRS:
     h(f(x,y)) -> f(f(a(),h(h(y))),x)
   Subterm Criterion Processor:
    simple projection:
     pi(h#) = 0
    problem:
     DPs:
      
     TRS:
      h(f(x,y)) -> f(f(a(),h(h(y))),x)
    Qed