YES

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

Proof:
 KBO Processor:
  weight function:
   w0 = 1
   w(a) = 1
   w(b) = 0
  precedence:
   b > a
  problem:
   
  Qed