*** 1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        add(0(),X) -> X
        add(s(),Y) -> s()
        dbl(0()) -> 0()
        dbl(s()) -> s()
        first(0(),X) -> nil()
        first(s(),cons(Y)) -> cons(Y)
        sqr(0()) -> 0()
        sqr(s()) -> s()
        terms(N) -> cons(recip(sqr(N)))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {add/2,dbl/1,first/2,sqr/1,terms/1} / {0/0,cons/1,nil/0,recip/1,s/0}
      Obligation:
        Full
        basic terms: {add,dbl,first,sqr,terms}/{0,cons,nil,recip,s}
    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:
        add(0(),X) -> X
        add(s(),Y) -> s()
        dbl(0()) -> 0()
        dbl(s()) -> s()
        first(0(),X) -> nil()
        first(s(),cons(Y)) -> cons(Y)
        sqr(0()) -> 0()
        sqr(s()) -> s()
        terms(N) -> cons(recip(sqr(N)))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {add/2,dbl/1,first/2,sqr/1,terms/1} / {0/0,cons/1,nil/0,recip/1,s/0}
      Obligation:
        Innermost
        basic terms: {add,dbl,first,sqr,terms}/{0,cons,nil,recip,s}
    Applied Processor:
      DependencyPairs {dpKind_ = DT}
    Proof:
      We add the following dependency tuples:
      
      Strict DPs
        add#(0(),X) -> c_1()
        add#(s(),Y) -> c_2()
        dbl#(0()) -> c_3()
        dbl#(s()) -> c_4()
        first#(0(),X) -> c_5()
        first#(s(),cons(Y)) -> c_6()
        sqr#(0()) -> c_7()
        sqr#(s()) -> c_8()
        terms#(N) -> c_9(sqr#(N))
      Weak DPs
        
      
      and mark the set of starting terms.
*** 1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        add#(0(),X) -> c_1()
        add#(s(),Y) -> c_2()
        dbl#(0()) -> c_3()
        dbl#(s()) -> c_4()
        first#(0(),X) -> c_5()
        first#(s(),cons(Y)) -> c_6()
        sqr#(0()) -> c_7()
        sqr#(s()) -> c_8()
        terms#(N) -> c_9(sqr#(N))
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        add(0(),X) -> X
        add(s(),Y) -> s()
        dbl(0()) -> 0()
        dbl(s()) -> s()
        first(0(),X) -> nil()
        first(s(),cons(Y)) -> cons(Y)
        sqr(0()) -> 0()
        sqr(s()) -> s()
        terms(N) -> cons(recip(sqr(N)))
      Signature:
        {add/2,dbl/1,first/2,sqr/1,terms/1,add#/2,dbl#/1,first#/2,sqr#/1,terms#/1} / {0/0,cons/1,nil/0,recip/1,s/0,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/0,c_7/0,c_8/0,c_9/1}
      Obligation:
        Innermost
        basic terms: {add#,dbl#,first#,sqr#,terms#}/{0,cons,nil,recip,s}
    Applied Processor:
      UsableRules
    Proof:
      We replace rewrite rules by usable rules:
        add#(0(),X) -> c_1()
        add#(s(),Y) -> c_2()
        dbl#(0()) -> c_3()
        dbl#(s()) -> c_4()
        first#(0(),X) -> c_5()
        first#(s(),cons(Y)) -> c_6()
        sqr#(0()) -> c_7()
        sqr#(s()) -> c_8()
        terms#(N) -> c_9(sqr#(N))
*** 1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        add#(0(),X) -> c_1()
        add#(s(),Y) -> c_2()
        dbl#(0()) -> c_3()
        dbl#(s()) -> c_4()
        first#(0(),X) -> c_5()
        first#(s(),cons(Y)) -> c_6()
        sqr#(0()) -> c_7()
        sqr#(s()) -> c_8()
        terms#(N) -> c_9(sqr#(N))
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {add/2,dbl/1,first/2,sqr/1,terms/1,add#/2,dbl#/1,first#/2,sqr#/1,terms#/1} / {0/0,cons/1,nil/0,recip/1,s/0,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/0,c_7/0,c_8/0,c_9/1}
      Obligation:
        Innermost
        basic terms: {add#,dbl#,first#,sqr#,terms#}/{0,cons,nil,recip,s}
    Applied Processor:
      Trivial
    Proof:
      Consider the dependency graph
        1:S:add#(0(),X) -> c_1()
           
        
        2:S:add#(s(),Y) -> c_2()
           
        
        3:S:dbl#(0()) -> c_3()
           
        
        4:S:dbl#(s()) -> c_4()
           
        
        5:S:first#(0(),X) -> c_5()
           
        
        6:S:first#(s(),cons(Y)) -> c_6()
           
        
        7:S:sqr#(0()) -> c_7()
           
        
        8:S:sqr#(s()) -> c_8()
           
        
        9:S:terms#(N) -> c_9(sqr#(N))
           -->_1 sqr#(s()) -> c_8():8
           -->_1 sqr#(0()) -> c_7():7
        
      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:
        {add/2,dbl/1,first/2,sqr/1,terms/1,add#/2,dbl#/1,first#/2,sqr#/1,terms#/1} / {0/0,cons/1,nil/0,recip/1,s/0,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/0,c_7/0,c_8/0,c_9/1}
      Obligation:
        Innermost
        basic terms: {add#,dbl#,first#,sqr#,terms#}/{0,cons,nil,recip,s}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).