YES

Problem:
 a(b(x1)) -> b(a(x1))
 b(a(x1)) -> a(c(b(x1)))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {4,1}
   transitions:
    a0(2) -> 3*
    a0(6) -> 4*
    a1(9) -> 10*
    c0(5) -> 6*
    b0(3) -> 1*
    b0(2) -> 5*
    f30() -> 2*
    c1(8) -> 9*
    b1(15) -> 16*
    b1(7) -> 8*
    16 -> 8*
    4 -> 8,5
    2 -> 7*
    1 -> 3*
    9 -> 15*
    10 -> 1*
  problem:
   
  Qed