MAYBE

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

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