*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        active(cons(X1,X2)) -> cons(active(X1),X2)
        active(f(X)) -> f(active(X))
        active(f(0())) -> mark(cons(0(),f(s(0()))))
        active(f(s(0()))) -> mark(f(p(s(0()))))
        active(p(X)) -> p(active(X))
        active(p(s(X))) -> mark(X)
        active(s(X)) -> s(active(X))
        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())
        proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
        proper(f(X)) -> f(proper(X))
        proper(p(X)) -> p(proper(X))
        proper(s(X)) -> s(proper(X))
        s(mark(X)) -> mark(s(X))
        s(ok(X)) -> ok(s(X))
        top(mark(X)) -> top(proper(X))
        top(ok(X)) -> top(active(X))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {active/1,cons/2,f/1,p/1,proper/1,s/1,top/1} / {0/0,mark/1,ok/1}
      Obligation:
        Full
        basic terms: {active,cons,f,p,proper,s,top}/{0,mark,ok}
    Applied Processor:
      Bounds {initialAutomaton = perSymbol, enrichment = match}
    Proof:
      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(5) -> 2
        active_0(6) -> 2
        active_1(1) -> 16
        active_1(5) -> 16
        active_1(6) -> 16
        active_2(14) -> 17
        cons_0(1,1) -> 3
        cons_0(1,5) -> 3
        cons_0(1,6) -> 3
        cons_0(5,1) -> 3
        cons_0(5,5) -> 3
        cons_0(5,6) -> 3
        cons_0(6,1) -> 3
        cons_0(6,5) -> 3
        cons_0(6,6) -> 3
        cons_1(1,1) -> 11
        cons_1(1,5) -> 11
        cons_1(1,6) -> 11
        cons_1(5,1) -> 11
        cons_1(5,5) -> 11
        cons_1(5,6) -> 11
        cons_1(6,1) -> 11
        cons_1(6,5) -> 11
        cons_1(6,6) -> 11
        f_0(1) -> 4
        f_0(5) -> 4
        f_0(6) -> 4
        f_1(1) -> 12
        f_1(5) -> 12
        f_1(6) -> 12
        mark_0(1) -> 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(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(5) -> 7
        p_0(6) -> 7
        p_1(1) -> 13
        p_1(5) -> 13
        p_1(6) -> 13
        proper_0(1) -> 8
        proper_0(5) -> 8
        proper_0(6) -> 8
        proper_1(1) -> 16
        proper_1(5) -> 16
        proper_1(6) -> 16
        s_0(1) -> 9
        s_0(5) -> 9
        s_0(6) -> 9
        s_1(1) -> 15
        s_1(5) -> 15
        s_1(6) -> 15
        top_0(1) -> 10
        top_0(5) -> 10
        top_0(6) -> 10
        top_1(16) -> 10
        top_2(17) -> 10
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        active(cons(X1,X2)) -> cons(active(X1),X2)
        active(f(X)) -> f(active(X))
        active(f(0())) -> mark(cons(0(),f(s(0()))))
        active(f(s(0()))) -> mark(f(p(s(0()))))
        active(p(X)) -> p(active(X))
        active(p(s(X))) -> mark(X)
        active(s(X)) -> s(active(X))
        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())
        proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
        proper(f(X)) -> f(proper(X))
        proper(p(X)) -> p(proper(X))
        proper(s(X)) -> s(proper(X))
        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:
        {active/1,cons/2,f/1,p/1,proper/1,s/1,top/1} / {0/0,mark/1,ok/1}
      Obligation:
        Full
        basic terms: {active,cons,f,p,proper,s,top}/{0,mark,ok}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).