YES

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

Proof:
 String Reversal Processor:
  a(b(x1)) -> b(a(x1))
  a(a(a(x1))) -> b(a(a(b(x1))))
  b(b(b(b(x1)))) -> a(x1)
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(a) = 2
    w(b) = 1
   status function:
    st(b) = st(a) = [0]
   precedence:
    a > b
   problem:
    
   Qed