YES

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(a(),a())) -> f#(x,a())
   f#(x,f(a(),a())) -> f#(f(a(),a()),a())
   f#(x,f(a(),a())) -> f#(f(f(a(),a()),a()),f(x,a()))
  TRS:
   f(x,f(a(),a())) -> f(f(f(a(),a()),a()),f(x,a()))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [f#](x0, x1) = [1 0 1 1]x0 + [0 1 1 0]x1 + [1],
    
                  [1 0 0 0]     [0]
                  [0 0 1 1]     [1]
    [f](x0, x1) = [1 0 0 0]x0 + [0]
                  [0 0 0 0]     [0],
    
          [0]
          [0]
    [a] = [1]
          [0]
   orientation:
    f#(x,f(a(),a())) = [1 0 1 1]x + [3] >= [1 0 1 1]x + [2] = f#(x,a())
    
    f#(x,f(a(),a())) = [1 0 1 1]x + [3] >= [2] = f#(f(a(),a()),a())
    
    f#(x,f(a(),a())) = [1 0 1 1]x + [3] >= [1 0 1 1]x + [2] = f#(f(f(a(),a()),a()),f(x,a()))
    
                      [1 0 0 0]    [0]    [0]                                
                      [0 0 1 1]    [1]    [1]                                
    f(x,f(a(),a())) = [1 0 0 0]x + [0] >= [0] = f(f(f(a(),a()),a()),f(x,a()))
                      [0 0 0 0]    [0]    [0]                                
   problem:
    DPs:
     
    TRS:
     f(x,f(a(),a())) -> f(f(f(a(),a()),a()),f(x,a()))
   Qed