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:
  { g(0()) -> 0()
  , g(s(x)) -> f(g(x))
  , f(0()) -> 0() }
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:
  { g(0()) -> 0()
  , g(s(x)) -> f(g(x))
  , f(0()) -> 0() }

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(g) = {}, safe(0) = {}, safe(s) = {1}, safe(f) = {1}
  
  and precedence
  
   g > f .
  
  Following symbols are considered recursive:
  
   {g}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
       g(0();) > 0()       
                           
    g(s(; x);) > f(; g(x;))
                           
      f(; 0()) > 0()       
                           

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:
  { g(0()) -> 0()
  , g(s(x)) -> f(g(x))
  , f(0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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