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 use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs:
  { app(nil(), xs) -> nil()
  , app(cons(x, xs), ys) -> cons(x, app(xs, ys)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(app) = {}, safe(nil) = {}, safe(cons) = {1, 2}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {app}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
             app(nil(),  xs;) > nil()                    
                                                         
    app(cons(; x,  xs),  ys;) > cons(; x,  app(xs,  ys;))
                                                         

We return to the main proof.

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

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

Empty rules are trivially bounded

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