MAYBE

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

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