YES

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

Proof:
 String Reversal Processor:
  a(a(x)) -> b(b(x))
  a(b(b(x))) -> b(b(a(x)))
  KBO Processor:
   weight function:
    w0 = 1
    w(b) = w(a) = 1
   precedence:
    a > b
   problem:
    
   Qed