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