YES

Problem:
 A(b(x1)) -> b(a(B(A(x1))))
 B(a(x1)) -> a(b(A(B(x1))))
 A(a(x1)) -> x1
 B(b(x1)) -> x1

Proof:
 Bounds Processor:
  bound: 0
  enrichment: match
  automaton:
   final states: {2,6,1}
   transitions:
    b0(5) -> 1*
    b0(8) -> 9*
    B0(3) -> 4*
    B0(2) -> 7*
    A0(2) -> 3*
    A0(7) -> 8*
    f40() -> 2*
    a0(4) -> 5*
    a0(9) -> 6*
    2 -> 4,8,7,3
    6 -> 4,7
    1 -> 8,3
    5 -> 4*
    9 -> 8*
  problem:
   
  Qed