MAYBE

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

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