MAYBE

Problem:
 a(b(x)) -> b(b(a(a(x))))

Proof:
 DP Processor:
  DPs:
   a#(b(x)) -> a#(x)
   a#(b(x)) -> a#(a(x))
  TRS:
   a(b(x)) -> b(b(a(a(x))))
  SCC Processor:
   #sccs: 1
   #rules: 2
   #arcs: 4/4
   DPs:
    a#(b(x)) -> a#(x)
    a#(b(x)) -> a#(a(x))
   TRS:
    a(b(x)) -> b(b(a(a(x))))
   Open