MAYBE

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

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