YES

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

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