MAYBE

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

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