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)
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
    [h#](x0) = [0  -&]x0 + [0],
    
          [0]
    [a] = [1],
    
              [-& 0 ]     [0]
    [h](x0) = [1  -&]x0 + [2],
    
                  [-& -&]     [0  1 ]     [2]
    [f](x0, x1) = [0  1 ]x0 + [-& -&]x1 + [2]
   orientation:
    h#(f(x,y)) = [0 1]y + [2] >= [0  -&]y + [0] = h#(y)
    
    h#(f(x,y)) = [0 1]y + [2] >= [-& 0 ]y + [0] = h#(h(y))
    
                [0  1 ]    [-& -&]    [2]    [0  1 ]    [-& -&]    [2]                      
    h(f(x,y)) = [-& -&]x + [1  2 ]y + [3] >= [-& -&]x + [1  2 ]y + [3] = f(f(a(),h(h(y))),x)
   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