YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
   a#(a(b(x1))) -> a#(x1)
  TRS:
   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))
  KBO Processor:
   weight function:
    w0 = 1
    w(a#) = w(a) = w(b) = 1
   precedence:
    a# > a > b
   problem:
    DPs:
     
    TRS:
     
   Qed