MAYBE

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

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