*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
        dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
        main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
        revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
        revApp#2(Nil(),x16) -> x16
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
      Obligation:
        Innermost
        basic terms: {dfsAcc#3,main,revApp#2}/{Cons,Leaf,Nil,Node}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 2.
      The enriched problem is compatible with follwoing automaton.
        Cons_0(2,2) -> 1
        Cons_0(2,2) -> 2
        Cons_1(2,1) -> 1
        Cons_1(2,2) -> 1
        Cons_1(2,2) -> 3
        Cons_1(2,3) -> 1
        Cons_1(2,3) -> 3
        Cons_1(2,4) -> 4
        Cons_1(2,6) -> 4
        Cons_2(2,5) -> 1
        Cons_2(2,5) -> 7
        Cons_2(2,7) -> 1
        Cons_2(2,7) -> 7
        Leaf_0(2) -> 1
        Leaf_0(2) -> 2
        Nil_0() -> 1
        Nil_0() -> 2
        Nil_1() -> 5
        Nil_1() -> 6
        Node_0(2,2) -> 1
        Node_0(2,2) -> 2
        dfsAcc#3_0(2,2) -> 1
        dfsAcc#3_1(2,2) -> 3
        dfsAcc#3_1(2,3) -> 1
        dfsAcc#3_1(2,3) -> 3
        dfsAcc#3_1(2,4) -> 4
        dfsAcc#3_1(2,6) -> 4
        main_0(2) -> 1
        revApp#2_0(2,2) -> 1
        revApp#2_1(2,1) -> 1
        revApp#2_1(4,5) -> 1
        revApp#2_2(4,7) -> 1
        revApp#2_2(6,7) -> 1
        2 -> 1
        7 -> 1
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        dfsAcc#3(Leaf(x8),x16) -> Cons(x8,x16)
        dfsAcc#3(Node(x6,x4),x2) -> dfsAcc#3(x4,dfsAcc#3(x6,x2))
        main(x1) -> revApp#2(dfsAcc#3(x1,Nil()),Nil())
        revApp#2(Cons(x6,x4),x2) -> revApp#2(x4,Cons(x6,x2))
        revApp#2(Nil(),x16) -> x16
      Signature:
        {dfsAcc#3/2,main/1,revApp#2/2} / {Cons/2,Leaf/1,Nil/0,Node/2}
      Obligation:
        Innermost
        basic terms: {dfsAcc#3,main,revApp#2}/{Cons,Leaf,Nil,Node}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).