MAYBE

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

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