YES

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

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