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:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sqr(s(x)), sum(x))
  , sum(s(x)) -> +(*(s(x), s(x)), sum(x))
  , sqr(x) -> *(x, x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x))
  , sqr^#(x) -> c_4() }

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:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x))
  , sqr^#(x) -> c_4() }
Strict Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sqr(s(x)), sum(x))
  , sum(s(x)) -> +(*(s(x), s(x)), sum(x))
  , sqr(x) -> *(x, x) }
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:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x))
  , sqr^#(x) -> c_4() }
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_2) = {1, 2}, Uargs(c_3) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

            [0] = [2]                  
                                       
        [s](x1) = [1] x1 + [0]         
                                       
    [sum^#](x1) = [0]                  
                                       
          [c_1] = [1]                  
                                       
  [c_2](x1, x2) = [1] x1 + [1] x2 + [0]
                                       
    [sqr^#](x1) = [1]                  
                                       
      [c_3](x1) = [1] x1 + [2]         
                                       
          [c_4] = [0]                  

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:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x)) }
Weak DPs: { sqr^#(x) -> c_4() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: sum^#(0()) -> c_1()
    , 2: sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
    , 3: sum^#(s(x)) -> c_3(sum^#(x))
    , 4: sqr^#(x) -> c_4() }

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

Strict DPs:
  { sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x)) }
Weak DPs:
  { sum^#(0()) -> c_1()
  , sqr^#(x) -> c_4() }
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.

{ sum^#(0()) -> c_1()
, sqr^#(x) -> c_4() }

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

Strict DPs:
  { sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x))
  , sum^#(s(x)) -> c_3(sum^#(x)) }
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:

  { sum^#(s(x)) -> c_2(sqr^#(s(x)), sum^#(x)) }

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

Strict DPs:
  { sum^#(s(x)) -> c_1(sum^#(x))
  , sum^#(s(x)) -> c_2(sum^#(x)) }
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(sum^#) = {}, safe(c_1) = {}, safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {sum^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(s) = [1], pi(sum^#) = [1], pi(c_1) = [1], pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {sum^#}

For your convenience, here are the satisfied ordering constraints:

  pi(sum^#(s(x))) = sum^#(s(; x);)   
                  > c_1(sum^#(x;);)  
                  = pi(c_1(sum^#(x)))
                                     
  pi(sum^#(s(x))) = sum^#(s(; x);)   
                  > c_2(sum^#(x;);)  
                  = pi(c_2(sum^#(x)))
                                     

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