YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(x1))) -> b#(b(a(x1)))
   a#(b(a(x1))) -> a#(b(b(a(x1))))
  TRS:
   a(b(a(x1))) -> a(b(b(a(x1))))
   b(b(b(x1))) -> b(b(x1))
  TDG Processor:
   DPs:
    a#(b(a(x1))) -> b#(b(a(x1)))
    a#(b(a(x1))) -> a#(b(b(a(x1))))
   TRS:
    a(b(a(x1))) -> a(b(b(a(x1))))
    b(b(b(x1))) -> b(b(x1))
   graph:
    a#(b(a(x1))) -> a#(b(b(a(x1)))) ->
    a#(b(a(x1))) -> a#(b(b(a(x1))))
    a#(b(a(x1))) -> a#(b(b(a(x1)))) -> a#(b(a(x1))) -> b#(b(a(x1)))
   CDG Processor:
    DPs:
     a#(b(a(x1))) -> b#(b(a(x1)))
     a#(b(a(x1))) -> a#(b(b(a(x1))))
    TRS:
     a(b(a(x1))) -> a(b(b(a(x1))))
     b(b(b(x1))) -> b(b(x1))
    graph:
     
    Qed