MAYBE

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

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