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:
  { norm(nil()) -> 0()
  , norm(g(x, y)) -> s(norm(x))
  , f(x, nil()) -> g(nil(), x)
  , f(x, g(y, z)) -> g(f(x, y), z)
  , rem(nil(), y) -> nil()
  , rem(g(x, y), 0()) -> g(x, y)
  , rem(g(x, y), s(z)) -> rem(x, z) }
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:
  { norm(nil()) -> 0()
  , norm(g(x, y)) -> s(norm(x))
  , f(x, nil()) -> g(nil(), x)
  , f(x, g(y, z)) -> g(f(x, y), z)
  , rem(nil(), y) -> nil()
  , rem(g(x, y), 0()) -> g(x, y)
  , rem(g(x, y), s(z)) -> rem(x, z) }

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(norm) = {}, safe(nil) = {}, safe(0) = {}, safe(g) = {1, 2},
   safe(s) = {1}, safe(f) = {}, safe(rem) = {2}
  
  and precedence
  
   norm ~ f, norm ~ rem, f ~ rem .
  
  Following symbols are considered recursive:
  
   {norm, f, rem}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
               norm(nil();) > 0()               
                                                
          norm(g(; x,  y);) > s(; norm(x;))     
                                                
              f(x,  nil();) > g(; nil(),  x)    
                                                
         f(x,  g(; y,  z);) > g(; f(x,  y;),  z)
                                                
              rem(nil(); y) > nil()             
                                                
       rem(g(; x,  y); 0()) > g(; x,  y)        
                                                
    rem(g(; x,  y); s(; z)) > rem(x; z)         
                                                

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

Empty rules are trivially bounded

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