MAYBE

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

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