YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 0 1]     [1 0 1]     [0]
   [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1]
                 [0 1 1]     [0 0 1]     [0],
   
         [0]
   [a] = [0]
         [0]
  orientation:
                            [1 0 1]    [1]    [1 0 1]    [0]                         
   f(x,f(f(a(),a()),a())) = [0 0 0]x + [1] >= [0 0 0]x + [1] = f(f(a(),f(a(),a())),x)
                            [0 1 1]    [1]    [0 0 1]    [1]                         
  problem:
   
  Qed