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)
  EDG 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)
   graph:
    f#(x,f(a(),a())) -> f#(f(x,a()),x) ->
    f#(x,f(a(),a())) -> f#(x,a())
    f#(x,f(a(),a())) -> f#(f(x,a()),x) -> f#(x,f(a(),a())) -> f#(f(x,a()),x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     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) = [0 1 1 0]x0 + [0 1 0 0]x1,
      
                    [0 0 0 0]     [1]
                    [0 0 1 0]     [0]
      [f](x0, x1) = [0 0 0 0]x0 + [0]
                    [0 0 1 1]     [0],
      
            [0]
            [0]
      [a] = [1]
            [0]
     orientation:
      f#(x,f(a(),a())) = [0 1 1 0]x + [1] >= [0 1 1 0]x = f#(f(x,a()),x)
      
                        [0 0 0 0]    [1]    [0 0 0 0]    [1]                
                        [0 0 1 0]    [0]    [0 0 0 0]    [0]                
      f(x,f(a(),a())) = [0 0 0 0]x + [0] >= [0 0 0 0]x + [0] = f(f(x,a()),x)
                        [0 0 1 1]    [0]    [0 0 1 1]    [0]                
     problem:
      DPs:
       
      TRS:
       f(x,f(a(),a())) -> f(f(x,a()),x)
     Qed