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)
  Matrix Interpretation Processor: dim=2
   
   usable rules:
    h(f(x,y)) -> f(f(a(),h(h(y))),x)
   interpretation:
    [h#](x0) = [2 0]x0,
    
          [0]
    [a] = [0],
    
              [0 1]  
    [h](x0) = [3 0]x0,
    
                  [0 0]     [1 1]     [1]
    [f](x0, x1) = [1 1]x0 + [0 0]x1 + [1]
   orientation:
    h#(f(x,y)) = [2 2]y + [2] >= [2 0]y = h#(y)
    
    h#(f(x,y)) = [2 2]y + [2] >= [0 2]y = h#(h(y))
    
                [1 1]    [0 0]    [1]    [1 1]    [0 0]    [1]                      
    h(f(x,y)) = [0 0]x + [3 3]y + [3] >= [0 0]x + [3 3]y + [3] = f(f(a(),h(h(y))),x)
   problem:
    DPs:
     
    TRS:
     h(f(x,y)) -> f(f(a(),h(h(y))),x)
   Qed