MAYBE

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

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