*** 1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        activate(X) -> X
        activate(n__zeros()) -> zeros()
        tail(cons(X,XS)) -> activate(XS)
        zeros() -> cons(0(),n__zeros())
        zeros() -> n__zeros()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,tail/1,zeros/0} / {0/0,cons/2,n__zeros/0}
      Obligation:
        Full
        basic terms: {activate,tail,zeros}/{0,cons,n__zeros}
    Applied Processor:
      ToInnermost
    Proof:
      switch to innermost, as the system is overlay and right linear and does not contain weak rules
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        activate(X) -> X
        activate(n__zeros()) -> zeros()
        tail(cons(X,XS)) -> activate(XS)
        zeros() -> cons(0(),n__zeros())
        zeros() -> n__zeros()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,tail/1,zeros/0} / {0/0,cons/2,n__zeros/0}
      Obligation:
        Innermost
        basic terms: {activate,tail,zeros}/{0,cons,n__zeros}
    Applied Processor:
      DependencyPairs {dpKind_ = DT}
    Proof:
      We add the following dependency tuples:
      
      Strict DPs
        activate#(X) -> c_1()
        activate#(n__zeros()) -> c_2(zeros#())
        tail#(cons(X,XS)) -> c_3(activate#(XS))
        zeros#() -> c_4()
        zeros#() -> c_5()
      Weak DPs
        
      
      and mark the set of starting terms.
*** 1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        activate#(X) -> c_1()
        activate#(n__zeros()) -> c_2(zeros#())
        tail#(cons(X,XS)) -> c_3(activate#(XS))
        zeros#() -> c_4()
        zeros#() -> c_5()
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        activate(X) -> X
        activate(n__zeros()) -> zeros()
        tail(cons(X,XS)) -> activate(XS)
        zeros() -> cons(0(),n__zeros())
        zeros() -> n__zeros()
      Signature:
        {activate/1,tail/1,zeros/0,activate#/1,tail#/1,zeros#/0} / {0/0,cons/2,n__zeros/0,c_1/0,c_2/1,c_3/1,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {activate#,tail#,zeros#}/{0,cons,n__zeros}
    Applied Processor:
      UsableRules
    Proof:
      We replace rewrite rules by usable rules:
        activate#(X) -> c_1()
        activate#(n__zeros()) -> c_2(zeros#())
        tail#(cons(X,XS)) -> c_3(activate#(XS))
        zeros#() -> c_4()
        zeros#() -> c_5()
*** 1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        activate#(X) -> c_1()
        activate#(n__zeros()) -> c_2(zeros#())
        tail#(cons(X,XS)) -> c_3(activate#(XS))
        zeros#() -> c_4()
        zeros#() -> c_5()
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,tail/1,zeros/0,activate#/1,tail#/1,zeros#/0} / {0/0,cons/2,n__zeros/0,c_1/0,c_2/1,c_3/1,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {activate#,tail#,zeros#}/{0,cons,n__zeros}
    Applied Processor:
      Trivial
    Proof:
      Consider the dependency graph
        1:S:activate#(X) -> c_1()
           
        
        2:S:activate#(n__zeros()) -> c_2(zeros#())
           -->_1 zeros#() -> c_5():5
           -->_1 zeros#() -> c_4():4
        
        3:S:tail#(cons(X,XS)) -> c_3(activate#(XS))
           -->_1 activate#(n__zeros()) -> c_2(zeros#()):2
           -->_1 activate#(X) -> c_1():1
        
        4:S:zeros#() -> c_4()
           
        
        5:S:zeros#() -> c_5()
           
        
      The dependency graph contains no loops, we remove all dependency pairs.
*** 1.1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,tail/1,zeros/0,activate#/1,tail#/1,zeros#/0} / {0/0,cons/2,n__zeros/0,c_1/0,c_2/1,c_3/1,c_4/0,c_5/0}
      Obligation:
        Innermost
        basic terms: {activate#,tail#,zeros#}/{0,cons,n__zeros}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).