*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        anchored(Cons(x,xs),y) -> anchored(xs,Cons(Cons(Nil(),Nil()),y))
        anchored(Nil(),y) -> y
        goal(x,y) -> anchored(x,y)
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {anchored/2,goal/2} / {Cons/2,Nil/0}
      Obligation:
        Innermost
        basic terms: {anchored,goal}/{Cons,Nil}
    Applied Processor:
      Bounds {initialAutomaton = perSymbol, enrichment = match}
    Proof:
      The problem is match-bounded by 1.
      The enriched problem is compatible with follwoing automaton.
        Cons_0(1,1) -> 1
        Cons_0(1,1) -> 3
        Cons_0(1,1) -> 4
        Cons_0(1,2) -> 1
        Cons_0(1,2) -> 3
        Cons_0(1,2) -> 4
        Cons_0(2,1) -> 1
        Cons_0(2,1) -> 3
        Cons_0(2,1) -> 4
        Cons_0(2,2) -> 1
        Cons_0(2,2) -> 3
        Cons_0(2,2) -> 4
        Cons_1(6,1) -> 3
        Cons_1(6,1) -> 4
        Cons_1(6,1) -> 5
        Cons_1(6,2) -> 3
        Cons_1(6,2) -> 4
        Cons_1(6,2) -> 5
        Cons_1(6,5) -> 3
        Cons_1(6,5) -> 4
        Cons_1(6,5) -> 5
        Cons_1(7,8) -> 6
        Nil_0() -> 2
        Nil_0() -> 3
        Nil_0() -> 4
        Nil_1() -> 7
        Nil_1() -> 8
        anchored_0(1,1) -> 3
        anchored_0(1,2) -> 3
        anchored_0(2,1) -> 3
        anchored_0(2,2) -> 3
        anchored_1(1,1) -> 4
        anchored_1(1,2) -> 4
        anchored_1(1,5) -> 3
        anchored_1(1,5) -> 4
        anchored_1(2,1) -> 4
        anchored_1(2,2) -> 4
        anchored_1(2,5) -> 3
        anchored_1(2,5) -> 4
        goal_0(1,1) -> 4
        goal_0(1,2) -> 4
        goal_0(2,1) -> 4
        goal_0(2,2) -> 4
        1 -> 3
        1 -> 4
        2 -> 3
        2 -> 4
        5 -> 3
        5 -> 4
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        anchored(Cons(x,xs),y) -> anchored(xs,Cons(Cons(Nil(),Nil()),y))
        anchored(Nil(),y) -> y
        goal(x,y) -> anchored(x,y)
      Signature:
        {anchored/2,goal/2} / {Cons/2,Nil/0}
      Obligation:
        Innermost
        basic terms: {anchored,goal}/{Cons,Nil}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).