YES

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

Proof:
 String Reversal Processor:
  b(b(a(x1))) -> a(a(b(x1)))
  b(a(a(x1))) -> a(b(b(x1)))
  KBO Processor:
   weight function:
    w0 = 1
    w(a) = w(b) = 1
   precedence:
    b > a
   problem:
    
   Qed