MAYBE

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

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