*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        a__tail(X) -> tail(X)
        a__tail(cons(X,XS)) -> mark(XS)
        a__zeros() -> cons(0(),zeros())
        a__zeros() -> zeros()
        mark(0()) -> 0()
        mark(cons(X1,X2)) -> cons(mark(X1),X2)
        mark(tail(X)) -> a__tail(mark(X))
        mark(zeros()) -> a__zeros()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {a__tail/1,a__zeros/0,mark/1} / {0/0,cons/2,tail/1,zeros/0}
      Obligation:
        Innermost
        basic terms: {a__tail,a__zeros,mark}/{0,cons,tail,zeros}
    Applied Processor:
      Bounds {initialAutomaton = perSymbol, enrichment = match}
    Proof:
      The problem is match-bounded by 4.
      The enriched problem is compatible with follwoing automaton.
        0_0() -> 1
        0_1() -> 2
        0_1() -> 5
        0_1() -> 8
        0_2() -> 10
        0_4() -> 12
        a__tail_0(1) -> 2
        a__tail_0(4) -> 2
        a__tail_0(6) -> 2
        a__tail_0(7) -> 2
        a__tail_1(2) -> 2
        a__tail_1(2) -> 5
        a__zeros_0() -> 3
        a__zeros_1() -> 2
        a__zeros_1() -> 5
        a__zeros_3() -> 2
        a__zeros_3() -> 5
        cons_0(1,1) -> 4
        cons_0(1,4) -> 4
        cons_0(1,6) -> 4
        cons_0(1,7) -> 4
        cons_0(4,1) -> 4
        cons_0(4,4) -> 4
        cons_0(4,6) -> 4
        cons_0(4,7) -> 4
        cons_0(6,1) -> 4
        cons_0(6,4) -> 4
        cons_0(6,6) -> 4
        cons_0(6,7) -> 4
        cons_0(7,1) -> 4
        cons_0(7,4) -> 4
        cons_0(7,6) -> 4
        cons_0(7,7) -> 4
        cons_1(2,1) -> 2
        cons_1(2,1) -> 5
        cons_1(2,4) -> 2
        cons_1(2,4) -> 5
        cons_1(2,6) -> 2
        cons_1(2,6) -> 5
        cons_1(2,7) -> 2
        cons_1(2,7) -> 5
        cons_1(8,9) -> 3
        cons_2(10,11) -> 2
        cons_2(10,11) -> 5
        cons_4(12,13) -> 2
        cons_4(12,13) -> 5
        mark_0(1) -> 5
        mark_0(4) -> 5
        mark_0(6) -> 5
        mark_0(7) -> 5
        mark_1(1) -> 2
        mark_1(4) -> 2
        mark_1(6) -> 2
        mark_1(7) -> 2
        mark_2(1) -> 2
        mark_2(1) -> 5
        mark_2(4) -> 2
        mark_2(4) -> 5
        mark_2(6) -> 2
        mark_2(6) -> 5
        mark_2(7) -> 2
        mark_2(7) -> 5
        mark_2(11) -> 2
        mark_2(11) -> 5
        mark_2(13) -> 2
        mark_2(13) -> 5
        tail_0(1) -> 6
        tail_0(4) -> 6
        tail_0(6) -> 6
        tail_0(7) -> 6
        tail_1(1) -> 2
        tail_1(4) -> 2
        tail_1(6) -> 2
        tail_1(7) -> 2
        tail_2(2) -> 2
        tail_2(2) -> 5
        zeros_0() -> 7
        zeros_1() -> 3
        zeros_1() -> 9
        zeros_2() -> 2
        zeros_2() -> 5
        zeros_2() -> 11
        zeros_4() -> 2
        zeros_4() -> 5
        zeros_4() -> 13
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        a__tail(X) -> tail(X)
        a__tail(cons(X,XS)) -> mark(XS)
        a__zeros() -> cons(0(),zeros())
        a__zeros() -> zeros()
        mark(0()) -> 0()
        mark(cons(X1,X2)) -> cons(mark(X1),X2)
        mark(tail(X)) -> a__tail(mark(X))
        mark(zeros()) -> a__zeros()
      Signature:
        {a__tail/1,a__zeros/0,mark/1} / {0/0,cons/2,tail/1,zeros/0}
      Obligation:
        Innermost
        basic terms: {a__tail,a__zeros,mark}/{0,cons,tail,zeros}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).