MAYBE

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

Proof:
 DP Processor:
  DPs:
   b#(a(a(x1))) -> b#(x1)
   b#(a(a(x1))) -> b#(a(b(x1)))
  TRS:
   b(b(x1)) -> a(a(a(x1)))
   b(a(b(x1))) -> a(x1)
   b(a(a(x1))) -> b(a(b(x1)))
  SCC Processor:
   #sccs: 1
   #rules: 2
   #arcs: 4/4
   DPs:
    b#(a(a(x1))) -> b#(x1)
    b#(a(a(x1))) -> b#(a(b(x1)))
   TRS:
    b(b(x1)) -> a(a(a(x1)))
    b(a(b(x1))) -> a(x1)
    b(a(a(x1))) -> b(a(b(x1)))
   Open