YES

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
   f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
  TRS:
   f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
  EDG Processor:
   DPs:
    f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
   TRS:
    f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
   graph:
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x) ->
    f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x) ->
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
    TRS:
     f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
    Bounds Processor:
     bound: 1
     enrichment: match-dp
     automaton:
      final states: {4}
      transitions:
       f{#,1}(9,6) -> 4*
       f1(7,7) -> 8*
       f1(7,8) -> 9*
       a1() -> 7*
       f{#,0}(6,3) -> 4*
       f0(1,2) -> 2*
       f0(2,1) -> 2*
       f0(1,1) -> 5,2
       f0(1,5) -> 6*
       f0(2,2) -> 2*
       a0() -> 1*
       1 -> 3*
       2 -> 3*
     problem:
      DPs:
       
      TRS:
       f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
     Qed