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: 1
   enrichment: match
   automaton:
    final states: {8,7,1}
    transitions:
     a{2,1}(2,16) -> 17*
     a{2,1}(2,3) -> 10*
     a{2,1}(3,2) -> 6*
     f1(18,15) -> 6*
     f1(17,15) -> 18*
     f1(3,16) -> 17*
     a{1,1}(15) -> 16*
     a1() -> 15*
     f40() -> 4*
     f0(3,3) -> 10*
     f0(6,2) -> 1*
     f0(4,3) -> 5*
     f0(10,2) -> 5*
     f0(5,2) -> 6*
     a{1,0}(2) -> 3*
     a{1,0}(4) -> 8*
     a{1,0}(3) -> 5*
     a0() -> 2*
     a{2,0}(4,4) -> 7*
     a{2,0}(2,3) -> 10*
     a{2,0}(3,2) -> 6*
     a{2,0}(4,3) -> 5*
     1 -> 7*
   problem:
    
   Qed