NO

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

Proof:
 String Reversal Processor:
  a(a(a(x1))) -> b(b(a(x1)))
  b(a(b(x1))) -> a(b(a(x1)))
  Unfolding Processor:
   loop length: 6
   terms:
    a(a(a(b(a(a(x774))))))
    b(b(a(b(a(a(x774))))))
    b(a(b(a(a(a(x774))))))
    a(b(a(a(a(a(x774))))))
    a(b(a(b(b(a(x774))))))
    a(a(b(a(b(a(x774))))))
   context: []
   substitution:
    x774 -> x774
   Qed