MAYBE

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

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