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())
  Restore Modifier:
   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())
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 4/4
    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())
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [f#](x0, x1) = x0,
      
      [f](x0, x1) = 0,
      
      [a] = 1
     orientation:
      f#(a(),f(f(a(),x),a())) = 1 >= 1 = f#(a(),f(a(),x))
      
      f#(a(),f(f(a(),x),a())) = 1 >= 0 = f#(f(a(),f(a(),x)),a())
      
      f(a(),f(f(a(),x),a())) = 0 >= 0 = f(f(a(),f(a(),x)),a())
     problem:
      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:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x1,
       
       [f](x0, x1) = x0 + x1 + 1,
       
       [a] = 0
      orientation:
       f#(a(),f(f(a(),x),a())) = x + 2 >= x + 1 = f#(a(),f(a(),x))
       
       f(a(),f(f(a(),x),a())) = x + 3 >= x + 3 = 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