* Step 1: Bounds WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            cons(mark(X1),X2) -> mark(cons(X1,X2))
            cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
            f(mark(X)) -> mark(f(X))
            f(ok(X)) -> ok(f(X))
            p(mark(X)) -> mark(p(X))
            p(ok(X)) -> ok(p(X))
            proper(0()) -> ok(0())
            s(mark(X)) -> mark(s(X))
            s(ok(X)) -> ok(s(X))
            top(mark(X)) -> top(proper(X))
            top(ok(X)) -> top(active(X))
        - Signature:
            {cons/2,f/1,p/1,proper/1,s/1,top/1} / {0/0,active/1,mark/1,ok/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {cons,f,p,proper,s,top} and constructors {0,active,mark
            ,ok}
    + Applied Processor:
        Bounds {initialAutomaton = perSymbol, enrichment = match}
    + Details:
        The problem is match-bounded by 2.
        The enriched problem is compatible with follwoing automaton.
          0_0() -> 1
          0_1() -> 14
          active_0(1) -> 2
          active_0(2) -> 2
          active_0(5) -> 2
          active_0(6) -> 2
          active_1(1) -> 16
          active_1(2) -> 16
          active_1(5) -> 16
          active_1(6) -> 16
          active_2(14) -> 17
          cons_0(1,1) -> 3
          cons_0(1,2) -> 3
          cons_0(1,5) -> 3
          cons_0(1,6) -> 3
          cons_0(2,1) -> 3
          cons_0(2,2) -> 3
          cons_0(2,5) -> 3
          cons_0(2,6) -> 3
          cons_0(5,1) -> 3
          cons_0(5,2) -> 3
          cons_0(5,5) -> 3
          cons_0(5,6) -> 3
          cons_0(6,1) -> 3
          cons_0(6,2) -> 3
          cons_0(6,5) -> 3
          cons_0(6,6) -> 3
          cons_1(1,1) -> 11
          cons_1(1,2) -> 11
          cons_1(1,5) -> 11
          cons_1(1,6) -> 11
          cons_1(2,1) -> 11
          cons_1(2,2) -> 11
          cons_1(2,5) -> 11
          cons_1(2,6) -> 11
          cons_1(5,1) -> 11
          cons_1(5,2) -> 11
          cons_1(5,5) -> 11
          cons_1(5,6) -> 11
          cons_1(6,1) -> 11
          cons_1(6,2) -> 11
          cons_1(6,5) -> 11
          cons_1(6,6) -> 11
          f_0(1) -> 4
          f_0(2) -> 4
          f_0(5) -> 4
          f_0(6) -> 4
          f_1(1) -> 12
          f_1(2) -> 12
          f_1(5) -> 12
          f_1(6) -> 12
          mark_0(1) -> 5
          mark_0(2) -> 5
          mark_0(5) -> 5
          mark_0(6) -> 5
          mark_1(11) -> 3
          mark_1(11) -> 11
          mark_1(12) -> 4
          mark_1(12) -> 12
          mark_1(13) -> 7
          mark_1(13) -> 13
          mark_1(15) -> 9
          mark_1(15) -> 15
          ok_0(1) -> 6
          ok_0(2) -> 6
          ok_0(5) -> 6
          ok_0(6) -> 6
          ok_1(11) -> 3
          ok_1(11) -> 11
          ok_1(12) -> 4
          ok_1(12) -> 12
          ok_1(13) -> 7
          ok_1(13) -> 13
          ok_1(14) -> 8
          ok_1(14) -> 16
          ok_1(15) -> 9
          ok_1(15) -> 15
          p_0(1) -> 7
          p_0(2) -> 7
          p_0(5) -> 7
          p_0(6) -> 7
          p_1(1) -> 13
          p_1(2) -> 13
          p_1(5) -> 13
          p_1(6) -> 13
          proper_0(1) -> 8
          proper_0(2) -> 8
          proper_0(5) -> 8
          proper_0(6) -> 8
          proper_1(1) -> 16
          proper_1(2) -> 16
          proper_1(5) -> 16
          proper_1(6) -> 16
          s_0(1) -> 9
          s_0(2) -> 9
          s_0(5) -> 9
          s_0(6) -> 9
          s_1(1) -> 15
          s_1(2) -> 15
          s_1(5) -> 15
          s_1(6) -> 15
          top_0(1) -> 10
          top_0(2) -> 10
          top_0(5) -> 10
          top_0(6) -> 10
          top_1(16) -> 10
          top_2(17) -> 10
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            cons(mark(X1),X2) -> mark(cons(X1,X2))
            cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
            f(mark(X)) -> mark(f(X))
            f(ok(X)) -> ok(f(X))
            p(mark(X)) -> mark(p(X))
            p(ok(X)) -> ok(p(X))
            proper(0()) -> ok(0())
            s(mark(X)) -> mark(s(X))
            s(ok(X)) -> ok(s(X))
            top(mark(X)) -> top(proper(X))
            top(ok(X)) -> top(active(X))
        - Signature:
            {cons/2,f/1,p/1,proper/1,s/1,top/1} / {0/0,active/1,mark/1,ok/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {cons,f,p,proper,s,top} and constructors {0,active,mark
            ,ok}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))