YES(O(1),O(n^1))

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

Strict Trs:
  { prime(0()) -> false()
  , prime(s(0())) -> false()
  , prime(s(s(x))) -> prime1(s(s(x)), s(x))
  , prime1(x, 0()) -> false()
  , prime1(x, s(0())) -> true()
  , prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
  , divp(x, y) -> =(rem(x, y), 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { prime^#(0()) -> c_1()
  , prime^#(s(0())) -> c_2()
  , prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, 0()) -> c_4()
  , prime1^#(x, s(0())) -> c_5()
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y)))
  , divp^#(x, y) -> c_7() }

and mark the set of starting terms.

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

Strict DPs:
  { prime^#(0()) -> c_1()
  , prime^#(s(0())) -> c_2()
  , prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, 0()) -> c_4()
  , prime1^#(x, s(0())) -> c_5()
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y)))
  , divp^#(x, y) -> c_7() }
Strict Trs:
  { prime(0()) -> false()
  , prime(s(0())) -> false()
  , prime(s(s(x))) -> prime1(s(s(x)), s(x))
  , prime1(x, 0()) -> false()
  , prime1(x, s(0())) -> true()
  , prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
  , divp(x, y) -> =(rem(x, y), 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^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),O(n^1)).

Strict DPs:
  { prime^#(0()) -> c_1()
  , prime^#(s(0())) -> c_2()
  , prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, 0()) -> c_4()
  , prime1^#(x, s(0())) -> c_5()
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y)))
  , divp^#(x, y) -> c_7() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

The following argument positions are usable:
  Uargs(c_3) = {1}, Uargs(c_6) = {1, 2}

TcT has computed following constructor-restricted matrix
interpretation.

                 [0] = [0]                  
                                            
             [s](x1) = [1]                  
                                            
       [prime^#](x1) = [1]                  
                                            
               [c_1] = [0]                  
                                            
               [c_2] = [0]                  
                                            
           [c_3](x1) = [1] x1 + [0]         
                                            
  [prime1^#](x1, x2) = [1] x2 + [0]         
                                            
               [c_4] = [1]                  
                                            
               [c_5] = [0]                  
                                            
       [c_6](x1, x2) = [1] x1 + [1] x2 + [0]
                                            
    [divp^#](x1, x2) = [0]                  
                                            
               [c_7] = [1]                  

This order satisfies following ordering constraints:


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(n^1)).

Strict DPs:
  { prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, 0()) -> c_4()
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y)))
  , divp^#(x, y) -> c_7() }
Weak DPs:
  { prime^#(0()) -> c_1()
  , prime^#(s(0())) -> c_2()
  , prime1^#(x, s(0())) -> c_5() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {2,4} by applications of
Pre({2,4}) = {3}. Here rules are labeled as follows:

  DPs:
    { 1: prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
    , 2: prime1^#(x, 0()) -> c_4()
    , 3: prime1^#(x, s(s(y))) ->
         c_6(divp^#(s(s(y)), x), prime1^#(x, s(y)))
    , 4: divp^#(x, y) -> c_7()
    , 5: prime^#(0()) -> c_1()
    , 6: prime^#(s(0())) -> c_2()
    , 7: prime1^#(x, s(0())) -> c_5() }

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

Strict DPs:
  { prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y))) }
Weak DPs:
  { prime^#(0()) -> c_1()
  , prime^#(s(0())) -> c_2()
  , prime1^#(x, 0()) -> c_4()
  , prime1^#(x, s(0())) -> c_5()
  , divp^#(x, y) -> c_7() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

{ prime^#(0()) -> c_1()
, prime^#(s(0())) -> c_2()
, prime1^#(x, 0()) -> c_4()
, prime1^#(x, s(0())) -> c_5()
, divp^#(x, y) -> c_7() }

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

Strict DPs:
  { prime^#(s(s(x))) -> c_3(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { prime1^#(x, s(s(y))) ->
    c_6(divp^#(s(s(y)), x), prime1^#(x, s(y))) }

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

Strict DPs:
  { prime^#(s(s(x))) -> c_1(prime1^#(s(s(x)), s(x)))
  , prime1^#(x, s(s(y))) -> c_2(prime1^#(x, s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: prime^#(s(s(x))) -> c_1(prime1^#(s(s(x)), s(x)))
     -->_1 prime1^#(x, s(s(y))) -> c_2(prime1^#(x, s(y))) :2
  
  2: prime1^#(x, s(s(y))) -> c_2(prime1^#(x, s(y)))
     -->_1 prime1^#(x, s(s(y))) -> c_2(prime1^#(x, s(y))) :2
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { prime^#(s(s(x))) -> c_1(prime1^#(s(s(x)), s(x))) }


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

Strict DPs: { prime1^#(x, s(s(y))) -> c_2(prime1^#(x, s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

 safe(s) = {1}, safe(prime1^#) = {1}, safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {prime1^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(s) = [1], pi(prime1^#) = [1, 2], pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {prime1^#}

For your convenience, here are the satisfied ordering constraints:

  pi(prime1^#(x, s(s(y)))) = prime1^#(s(; s(; y)); x)  
                           > c_2(prime1^#(s(; y); x);) 
                           = pi(c_2(prime1^#(x, s(y))))
                                                       

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