YES

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

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