YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(x)) -> b#(x)
   a#(a(x)) -> b#(b(x))
   b#(b(a(x))) -> b#(x)
   b#(b(a(x))) -> b#(b(x))
   b#(b(a(x))) -> a#(b(b(x)))
  TRS:
   a(a(x)) -> b(b(x))
   b(b(a(x))) -> a(b(b(x)))
  KBO Processor:
   weight function:
    w0 = 1
    w(b#) = w(a#) = w(a) = 1
    w(b) = 0
   precedence:
    b > b# ~ a# ~ a
   problem:
    DPs:
     
    TRS:
     
   Qed