MAYBE

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

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