YES

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

Proof:
 Uncurry Processor (mirror):
  a2(x,a()) -> a1(f(f(x,a1(a())),a()))
  f(a1(x3),x4) -> a2(x3,x4)
  f(a(),x4) -> a1(x4)
  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{1,1}(13) -> 5*
     a{2,1}(13,5) -> 5*
     a{2,1}(5,5) -> 5*
     f1(13,11) -> 12*
     f1(5,11) -> 12*
     f1(12,10) -> 13*
     a1() -> 10*
     a{2,2}(13,11) -> 12*
     a{2,2}(5,11) -> 12*
     a{2,2}(10,15) -> 16*
     a{2,2}(11,10) -> 13*
     a{2,2}(17,11) -> 12*
     a{2,0}(5,5) -> 5*
     a{1,2}(17) -> 13*
     a{1,2}(14) -> 15*
     a0() -> 5*
     f2(16,14) -> 17*
     f2(11,15) -> 16*
     a{1,0}(5) -> 5*
     a2() -> 14*
     f0(5,5) -> 5*
   problem:
    
   Qed