YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a(),a()),x) -> f#(a(),f(a(),a()))
   f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a())))
  TRS:
   f(f(a(),a()),x) -> f(x,f(a(),f(a(),a())))
  EDG Processor:
   DPs:
    f#(f(a(),a()),x) -> f#(a(),f(a(),a()))
    f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a())))
   TRS:
    f(f(a(),a()),x) -> f(x,f(a(),f(a(),a())))
   graph:
    f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a()))) ->
    f#(f(a(),a()),x) -> f#(a(),f(a(),a()))
    f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a()))) -> f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a())))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),a())))
    TRS:
     f(f(a(),a()),x) -> f(x,f(a(),f(a(),a())))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {5}
      transitions:
       f1(9,10) -> 11*
       f1(9,9) -> 10*
       f1(4,11) -> 4*
       f1(11,11) -> 4*
       a1() -> 9*
       f{#,1}(4,11) -> 5*
       f{#,1}(11,11) -> 5*
       f{#,0}(4,4) -> 5*
       f0(4,4) -> 4*
       a0() -> 4*
     problem:
      DPs:
       
      TRS:
       f(f(a(),a()),x) -> f(x,f(a(),f(a(),a())))
     Qed