MAYBE

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

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