*** 1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        filter(cons(X),0(),M) -> cons(0())
        filter(cons(X),s(N),M) -> cons(X)
        nats(N) -> cons(N)
        sieve(cons(0())) -> cons(0())
        sieve(cons(s(N))) -> cons(s(N))
        zprimes() -> sieve(nats(s(s(0()))))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {filter/3,nats/1,sieve/1,zprimes/0} / {0/0,cons/1,s/1}
      Obligation:
        Innermost
        basic terms: {filter,nats,sieve,zprimes}/{0,cons,s}
    Applied Processor:
      DependencyPairs {dpKind_ = DT}
    Proof:
      We add the following dependency tuples:
      
      Strict DPs
        filter#(cons(X),0(),M) -> c_1()
        filter#(cons(X),s(N),M) -> c_2()
        nats#(N) -> c_3()
        sieve#(cons(0())) -> c_4()
        sieve#(cons(s(N))) -> c_5()
        zprimes#() -> c_6(sieve#(nats(s(s(0())))),nats#(s(s(0()))))
      Weak DPs
        
      
      and mark the set of starting terms.
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        filter#(cons(X),0(),M) -> c_1()
        filter#(cons(X),s(N),M) -> c_2()
        nats#(N) -> c_3()
        sieve#(cons(0())) -> c_4()
        sieve#(cons(s(N))) -> c_5()
        zprimes#() -> c_6(sieve#(nats(s(s(0())))),nats#(s(s(0()))))
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        filter(cons(X),0(),M) -> cons(0())
        filter(cons(X),s(N),M) -> cons(X)
        nats(N) -> cons(N)
        sieve(cons(0())) -> cons(0())
        sieve(cons(s(N))) -> cons(s(N))
        zprimes() -> sieve(nats(s(s(0()))))
      Signature:
        {filter/3,nats/1,sieve/1,zprimes/0,filter#/3,nats#/1,sieve#/1,zprimes#/0} / {0/0,cons/1,s/1,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/2}
      Obligation:
        Innermost
        basic terms: {filter#,nats#,sieve#,zprimes#}/{0,cons,s}
    Applied Processor:
      UsableRules
    Proof:
      We replace rewrite rules by usable rules:
        nats(N) -> cons(N)
        filter#(cons(X),0(),M) -> c_1()
        filter#(cons(X),s(N),M) -> c_2()
        nats#(N) -> c_3()
        sieve#(cons(0())) -> c_4()
        sieve#(cons(s(N))) -> c_5()
        zprimes#() -> c_6(sieve#(nats(s(s(0())))),nats#(s(s(0()))))
*** 1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        filter#(cons(X),0(),M) -> c_1()
        filter#(cons(X),s(N),M) -> c_2()
        nats#(N) -> c_3()
        sieve#(cons(0())) -> c_4()
        sieve#(cons(s(N))) -> c_5()
        zprimes#() -> c_6(sieve#(nats(s(s(0())))),nats#(s(s(0()))))
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        nats(N) -> cons(N)
      Signature:
        {filter/3,nats/1,sieve/1,zprimes/0,filter#/3,nats#/1,sieve#/1,zprimes#/0} / {0/0,cons/1,s/1,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/2}
      Obligation:
        Innermost
        basic terms: {filter#,nats#,sieve#,zprimes#}/{0,cons,s}
    Applied Processor:
      Trivial
    Proof:
      Consider the dependency graph
        1:S:filter#(cons(X),0(),M) -> c_1()
           
        
        2:S:filter#(cons(X),s(N),M) -> c_2()
           
        
        3:S:nats#(N) -> c_3()
           
        
        4:S:sieve#(cons(0())) -> c_4()
           
        
        5:S:sieve#(cons(s(N))) -> c_5()
           
        
        6:S:zprimes#() -> c_6(sieve#(nats(s(s(0())))),nats#(s(s(0()))))
           -->_1 sieve#(cons(s(N))) -> c_5():5
           -->_1 sieve#(cons(0())) -> c_4():4
           -->_2 nats#(N) -> c_3():3
        
      The dependency graph contains no loops, we remove all dependency pairs.
*** 1.1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        nats(N) -> cons(N)
      Signature:
        {filter/3,nats/1,sieve/1,zprimes/0,filter#/3,nats#/1,sieve#/1,zprimes#/0} / {0/0,cons/1,s/1,c_1/0,c_2/0,c_3/0,c_4/0,c_5/0,c_6/2}
      Obligation:
        Innermost
        basic terms: {filter#,nats#,sieve#,zprimes#}/{0,cons,s}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).