MAYBE

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

Proof:
 DP Processor:
  DPs:
   a#(a(x1)) -> b#(x1)
   b#(a(b(x1))) -> b#(b(x1))
   b#(a(b(x1))) -> b#(b(b(x1)))
   b#(a(b(x1))) -> a#(b(b(b(x1))))
   b#(a(b(x1))) -> a#(a(b(b(b(x1)))))
  TRS:
   a(x1) -> x1
   a(a(x1)) -> b(x1)
   b(a(b(x1))) -> a(a(b(b(b(x1)))))
  Open