YES

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

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {4,1}
   transitions:
    a0(7) -> 4*
    a0(2) -> 3*
    a0(6) -> 7*
    a1(11) -> 12*
    a1(10) -> 11*
    c0(5) -> 6*
    b2(27) -> 28*
    c2(28) -> 29*
    b0(3) -> 1*
    b0(2) -> 5*
    f30() -> 2*
    c1(9) -> 10*
    a2(30) -> 31*
    a2(29) -> 30*
    b1(18) -> 19*
    b1(8) -> 9*
    19 -> 9*
    12 -> 1*
    4 -> 9,5
    11 -> 18*
    2 -> 8*
    1 -> 3*
    31 -> 19,9
    10 -> 27*
  problem:
   
  Qed