MAYBE

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

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