MAYBE

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

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