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