YES(?,O(n^1))

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

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {2}
   transitions:
    b1(5) -> 6*
    b1(7) -> 8*
    a0(1) -> 2*
    b0(1) -> 1*
    1 -> 5*
    6 -> 7,2
    8 -> 5*
  problem:
   
  Qed