MAYBE

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

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