MAYBE

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

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