YES

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

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