YES

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

Proof:
 Uncurry Processor:
  a2(x,a()) -> f(f(f(x,a()),a1(a())),a())
  f(a1(x1),x2) -> a2(x1,x2)
  f(a(),x2) -> a1(x2)
  Bounds Processor:
   bound: 3
   enrichment: match
   automaton:
    final states: {5}
    transitions:
     f3(20,19) -> 21*
     f3(21,18) -> 20,16
     f3(5,18) -> 20*
     a{1,1}(10) -> 12,11
     a{1,1}(5) -> 5*
     a{1,1}(14) -> 16*
     a{1,1}(18) -> 20*
     a3() -> 18*
     a{2,1}(5,5) -> 5*
     a{1,3}(18) -> 19*
     f1(12,11) -> 13*
     f1(13,10) -> 5*
     f1(5,10) -> 12*
     a1() -> 10*
     a{2,2}(18,19) -> 21*
     a{2,2}(10,11) -> 13*
     a{2,2}(14,15) -> 17*
     a{2,2}(5,10) -> 12*
     a{2,2}(5,14) -> 16*
     a{2,2}(5,18) -> 20*
     a{2,0}(5,5) -> 5*
     f2(17,14) -> 12*
     f2(5,14) -> 16*
     f2(16,15) -> 17*
     a0() -> 5*
     a2() -> 14*
     f0(5,5) -> 5*
     a{1,2}(14) -> 15*
     a{1,0}(5) -> 5*
   problem:
    
   Qed