YES(O(1),O(1))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Strict Trs:
  { filter(cons(X), 0(), M) -> cons(0())
  , filter(cons(X), s(N), M) -> cons(X)
  , sieve(cons(0())) -> cons(0())
  , sieve(cons(s(N))) -> cons(s(N))
  , nats(N) -> cons(N)
  , zprimes() -> sieve(nats(s(s(0())))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

We add following weak dependency pairs:

Strict DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , nats^#(N) -> c_5()
  , zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Strict DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , nats^#(N) -> c_5()
  , zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }
Strict Trs:
  { filter(cons(X), 0(), M) -> cons(0())
  , filter(cons(X), s(N), M) -> cons(X)
  , sieve(cons(0())) -> cons(0())
  , sieve(cons(s(N))) -> cons(s(N))
  , nats(N) -> cons(N)
  , zprimes() -> sieve(nats(s(s(0())))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

We replace rewrite rules by usable rules:

  Strict Usable Rules: { nats(N) -> cons(N) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Strict DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , nats^#(N) -> c_5()
  , zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }
Strict Trs: { nats(N) -> cons(N) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(sieve^#) = {1}, Uargs(c_6) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

              [cons](x1) = [0]                  
                                                
                     [0] = [0]                  
                                                
                 [s](x1) = [0]                  
                                                
              [nats](x1) = [2]                  
                                                
  [filter^#](x1, x2, x3) = [2] x1 + [2] x2 + [2]
                                                
                   [c_1] = [1]                  
                                                
                   [c_2] = [1]                  
                                                
           [sieve^#](x1) = [2] x1 + [0]         
                                                
                   [c_3] = [1]                  
                                                
                   [c_4] = [1]                  
                                                
            [nats^#](x1) = [1]                  
                                                
                   [c_5] = [0]                  
                                                
             [zprimes^#] = [1]                  
                                                
               [c_6](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

  [nats(N)] = [2]      
            > [0]      
            = [cons(N)]
                       

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(1)).

Strict DPs:
  { sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }
Weak DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , nats^#(N) -> c_5() }
Weak Trs: { nats(N) -> cons(N) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

We estimate the number of application of {1,2} by applications of
Pre({1,2}) = {3}. Here rules are labeled as follows:

  DPs:
    { 1: sieve^#(cons(0())) -> c_3()
    , 2: sieve^#(cons(s(N))) -> c_4()
    , 3: zprimes^#() -> c_6(sieve^#(nats(s(s(0())))))
    , 4: filter^#(cons(X), 0(), M) -> c_1()
    , 5: filter^#(cons(X), s(N), M) -> c_2()
    , 6: nats^#(N) -> c_5() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(1)).

Strict DPs: { zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }
Weak DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , nats^#(N) -> c_5() }
Weak Trs: { nats(N) -> cons(N) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

We estimate the number of application of {1} by applications of
Pre({1}) = {}. Here rules are labeled as follows:

  DPs:
    { 1: zprimes^#() -> c_6(sieve^#(nats(s(s(0())))))
    , 2: filter^#(cons(X), 0(), M) -> c_1()
    , 3: filter^#(cons(X), s(N), M) -> c_2()
    , 4: sieve^#(cons(0())) -> c_3()
    , 5: sieve^#(cons(s(N))) -> c_4()
    , 6: nats^#(N) -> c_5() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(1)).

Weak DPs:
  { filter^#(cons(X), 0(), M) -> c_1()
  , filter^#(cons(X), s(N), M) -> c_2()
  , sieve^#(cons(0())) -> c_3()
  , sieve^#(cons(s(N))) -> c_4()
  , nats^#(N) -> c_5()
  , zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }
Weak Trs: { nats(N) -> cons(N) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ filter^#(cons(X), 0(), M) -> c_1()
, filter^#(cons(X), s(N), M) -> c_2()
, sieve^#(cons(0())) -> c_3()
, sieve^#(cons(s(N))) -> c_4()
, nats^#(N) -> c_5()
, zprimes^#() -> c_6(sieve^#(nats(s(s(0()))))) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(1)).

Weak Trs: { nats(N) -> cons(N) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

No rule is usable, rules are removed from the input problem.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(1)).

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping


and precedence

 empty .

Following symbols are considered recursive:

 {}

The recursion depth is 0.

Further, following argument filtering is employed:

 empty

Usable defined function symbols are a subset of:

 {}

For your convenience, here are the satisfied ordering constraints:


Hurray, we answered YES(O(1),O(1))