YES(?,O(n^1))

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

Strict Trs:
  { f(x, 0()) -> s(0())
  , f(s(x), s(y)) -> s(f(x, y))
  , g(0(), x) -> g(f(x, x), x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { f^#(x, 0()) -> c_1()
  , f^#(s(x), s(y)) -> c_2(f^#(x, y))
  , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }

and mark the set of starting terms.

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

Strict DPs:
  { f^#(x, 0()) -> c_1()
  , f^#(s(x), s(y)) -> c_2(f^#(x, y))
  , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }
Weak Trs:
  { f(x, 0()) -> s(0())
  , f(s(x), s(y)) -> s(f(x, y))
  , g(0(), x) -> g(f(x, x), x) }
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: f^#(x, 0()) -> c_1()
    , 2: f^#(s(x), s(y)) -> c_2(f^#(x, y))
    , 3: g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }

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

Strict DPs:
  { f^#(s(x), s(y)) -> c_2(f^#(x, y))
  , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }
Weak DPs: { f^#(x, 0()) -> c_1() }
Weak Trs:
  { f(x, 0()) -> s(0())
  , f(s(x), s(y)) -> s(f(x, y))
  , g(0(), x) -> g(f(x, x), x) }
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.

{ f^#(x, 0()) -> c_1() }

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

Strict DPs:
  { f^#(s(x), s(y)) -> c_2(f^#(x, y))
  , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }
Weak Trs:
  { f(x, 0()) -> s(0())
  , f(s(x), s(y)) -> s(f(x, y))
  , g(0(), x) -> g(f(x, x), 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:

  { g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) }

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

Strict DPs:
  { f^#(s(x), s(y)) -> c_1(f^#(x, y))
  , g^#(0(), x) -> c_2(f^#(x, x)) }
Weak Trs:
  { f(x, 0()) -> s(0())
  , f(s(x), s(y)) -> s(f(x, y))
  , g(0(), x) -> g(f(x, x), x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,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(n^1)).

Strict DPs:
  { f^#(s(x), s(y)) -> c_1(f^#(x, y))
  , g^#(0(), x) -> c_2(f^#(x, x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: f^#(s(x), s(y)) -> c_1(f^#(x, y))
     -->_1 f^#(s(x), s(y)) -> c_1(f^#(x, y)) :1
  
  2: g^#(0(), x) -> c_2(f^#(x, x))
     -->_1 f^#(s(x), s(y)) -> c_1(f^#(x, y)) :1
  

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).

  { g^#(0(), x) -> c_2(f^#(x, x)) }


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

Strict DPs: { f^#(s(x), s(y)) -> c_1(f^#(x, 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(f^#) = {2}, safe(c_1) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {f^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(s) = [1], pi(f^#) = [1, 2], pi(c_1) = [1]

Usable defined function symbols are a subset of:

 {f^#}

For your convenience, here are the satisfied ordering constraints:

  pi(f^#(s(x), s(y))) = f^#(s(; x); s(; y))
                      > c_1(f^#(x; y);)    
                      = pi(c_1(f^#(x, y))) 
                                           

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