YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
   f#(a(),f(f(a(),x),a())) -> f#(f(a(),f(a(),x)),a())
  TRS:
   f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
  EDG Processor:
   DPs:
    f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
    f#(a(),f(f(a(),x),a())) -> f#(f(a(),f(a(),x)),a())
   TRS:
    f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
   graph:
    f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x)) ->
    f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
    f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x)) ->
    f#(a(),f(f(a(),x),a())) -> f#(f(a(),f(a(),x)),a())
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
    TRS:
     f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [f#](x0, x1) = x1,
      
      [f](x0, x1) = 2x0 + 2x1 + 1,
      
      [a] = 0
     orientation:
      f#(a(),f(f(a(),x),a())) = 4x + 3 >= 2x + 1 = f#(a(),f(a(),x))
      
      f(a(),f(f(a(),x),a())) = 8x + 7 >= 8x + 7 = f(f(a(),f(a(),x)),a())
     problem:
      DPs:
       
      TRS:
       f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
     Qed