YES

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

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