MAYBE

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

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