* Step 1: Bounds WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            add(X1,mark(X2)) -> mark(add(X1,X2))
            add(mark(X1),X2) -> mark(add(X1,X2))
            add(ok(X1),ok(X2)) -> ok(add(X1,X2))
            cons(mark(X1),X2) -> mark(cons(X1,X2))
            cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
            from(mark(X)) -> mark(from(X))
            from(ok(X)) -> ok(from(X))
            fst(X1,mark(X2)) -> mark(fst(X1,X2))
            fst(mark(X1),X2) -> mark(fst(X1,X2))
            fst(ok(X1),ok(X2)) -> ok(fst(X1,X2))
            len(mark(X)) -> mark(len(X))
            len(ok(X)) -> ok(len(X))
            proper(0()) -> ok(0())
            proper(nil()) -> ok(nil())
            s(ok(X)) -> ok(s(X))
            top(mark(X)) -> top(proper(X))
            top(ok(X)) -> top(active(X))
        - Signature:
            {add/2,cons/2,from/1,fst/2,len/1,proper/1,s/1,top/1} / {0/0,active/1,mark/1,nil/0,ok/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {add,cons,from,fst,len,proper,s,top} and constructors {0
            ,active,mark,nil,ok}
    + Applied Processor:
        Bounds {initialAutomaton = minimal, enrichment = match}
    + Details:
        The problem is match-bounded by 2.
        The enriched problem is compatible with follwoing automaton.
          0_0() -> 2
          0_1() -> 3
          active_0(2) -> 2
          active_1(2) -> 4
          active_2(3) -> 5
          add_0(2,2) -> 1
          add_1(2,2) -> 3
          cons_0(2,2) -> 1
          cons_1(2,2) -> 3
          from_0(2) -> 1
          from_1(2) -> 3
          fst_0(2,2) -> 1
          fst_1(2,2) -> 3
          len_0(2) -> 1
          len_1(2) -> 3
          mark_0(2) -> 2
          mark_1(3) -> 1
          mark_1(3) -> 3
          nil_0() -> 2
          nil_1() -> 3
          ok_0(2) -> 2
          ok_1(3) -> 1
          ok_1(3) -> 3
          ok_1(3) -> 4
          proper_0(2) -> 1
          proper_1(2) -> 4
          s_0(2) -> 1
          s_1(2) -> 3
          top_0(2) -> 1
          top_1(4) -> 1
          top_2(5) -> 1
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            add(X1,mark(X2)) -> mark(add(X1,X2))
            add(mark(X1),X2) -> mark(add(X1,X2))
            add(ok(X1),ok(X2)) -> ok(add(X1,X2))
            cons(mark(X1),X2) -> mark(cons(X1,X2))
            cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
            from(mark(X)) -> mark(from(X))
            from(ok(X)) -> ok(from(X))
            fst(X1,mark(X2)) -> mark(fst(X1,X2))
            fst(mark(X1),X2) -> mark(fst(X1,X2))
            fst(ok(X1),ok(X2)) -> ok(fst(X1,X2))
            len(mark(X)) -> mark(len(X))
            len(ok(X)) -> ok(len(X))
            proper(0()) -> ok(0())
            proper(nil()) -> ok(nil())
            s(ok(X)) -> ok(s(X))
            top(mark(X)) -> top(proper(X))
            top(ok(X)) -> top(active(X))
        - Signature:
            {add/2,cons/2,from/1,fst/2,len/1,proper/1,s/1,top/1} / {0/0,active/1,mark/1,nil/0,ok/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {add,cons,from,fst,len,proper,s,top} and constructors {0
            ,active,mark,nil,ok}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))