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:
  { f(x, g(x)) -> x
  , f(x, h(y)) -> f(h(x), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

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

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:
  { f^#(x, g(x)) -> c_1()
  , f^#(x, h(y)) -> c_2(f^#(h(x), y)) }
Strict Trs:
  { f(x, g(x)) -> x
  , f(x, h(y)) -> f(h(x), y) }
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:
  { f^#(x, g(x)) -> c_1()
  , f^#(x, h(y)) -> c_2(f^#(h(x), y)) }
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}

TcT has computed following constructor-restricted matrix
interpretation.

        [g](x1) = [2]         
                              
        [h](x1) = [1] x1 + [0]
                              
  [f^#](x1, x2) = [2] x1 + [1]
                              
          [c_1] = [0]         
                              
      [c_2](x1) = [1] x1 + [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: { f^#(x, h(y)) -> c_2(f^#(h(x), y)) }
Weak DPs: { f^#(x, g(x)) -> c_1() }
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, g(x)) -> c_1() }

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

Strict DPs: { f^#(x, h(y)) -> c_2(f^#(h(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(h) = {1}, safe(f^#) = {1}, safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {f^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(h) = [1], pi(f^#) = [1, 2], pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {f^#}

For your convenience, here are the satisfied ordering constraints:

  pi(f^#(x, h(y))) = f^#(h(; y); x)       
                   > c_2(f^#(y; h(; x));) 
                   = pi(c_2(f^#(h(x), y)))
                                          

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