YES

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

Proof:
 Uncurry Processor:
  a2(x,a()) -> f(f(x,a1(a1(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}(3,15) -> 16*
     a{2,1}(4,2) -> 1*
     a{2,1}(3,4) -> 6*
     f1(4,15) -> 16*
     f1(16,13) -> 1*
     a{1,1}(14) -> 15*
     a{1,1}(13) -> 14*
     a1() -> 13*
     f40() -> 5*
     f0(4,4) -> 6*
     f0(6,2) -> 1*
     f0(5,4) -> 6*
     a{1,0}(5) -> 8*
     a{1,0}(2) -> 3*
     a{1,0}(4) -> 6*
     a{1,0}(3) -> 4*
     a0() -> 2*
     a{2,0}(4,2) -> 1,7
     a{2,0}(5,5) -> 7*
     a{2,0}(3,4) -> 6*
     a{2,0}(5,4) -> 6*
     1 -> 7*
   problem:
    
   Qed