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:
  { +(0(), y) -> y
  , +(s(x), y) -> +(x, s(y))
  , +(s(x), y) -> s(+(x, y)) }
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:
  { +(0(), y) -> y
  , +(s(x), y) -> +(x, s(y))
  , +(s(x), y) -> s(+(x, y)) }

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

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

Empty rules are trivially bounded

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