MAYBE

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

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