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:
  { app(nil(), xs) -> nil()
  , app(cons(x, xs), ys) -> cons(x, app(xs, ys)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { app^#(nil(), xs) -> c_1()
  , app^#(cons(x, xs), ys) -> c_2(app^#(xs, ys)) }

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:
  { app^#(nil(), xs) -> c_1()
  , app^#(cons(x, xs), ys) -> c_2(app^#(xs, ys)) }
Strict Trs:
  { app(nil(), xs) -> nil()
  , app(cons(x, xs), ys) -> cons(x, app(xs, ys)) }
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:
  { app^#(nil(), xs) -> c_1()
  , app^#(cons(x, xs), ys) -> c_2(app^#(xs, ys)) }
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.

            [nil] = [2]         
                                
   [cons](x1, x2) = [1] x2 + [1]
                                
  [app^#](x1, x2) = [2] x2 + [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: { app^#(cons(x, xs), ys) -> c_2(app^#(xs, ys)) }
Weak DPs: { app^#(nil(), xs) -> 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.

{ app^#(nil(), xs) -> c_1() }

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

Strict DPs: { app^#(cons(x, xs), ys) -> c_2(app^#(xs, ys)) }
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(cons) = {1, 2}, safe(app^#) = {2}, safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {app^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(cons) = [1, 2], pi(app^#) = [1, 2], pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {app^#}

For your convenience, here are the satisfied ordering constraints:

  pi(app^#(cons(x, xs), ys)) = app^#(cons(; x,  xs); ys)
                             > c_2(app^#(xs; ys);)      
                             = pi(c_2(app^#(xs, ys)))   
                                                        

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