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

Strict Trs:
  { not(true()) -> false()
  , not(false()) -> true()
  , odd(0()) -> false()
  , odd(s(x)) -> not(odd(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y))
  , +(s(x), y) -> s(+(x, y)) }
Obligation:
  runtime complexity
Answer:
  YES(?,O(n^1))

The input is overlay and right-linear. Switching to innermost
rewriting.

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

Strict Trs:
  { not(true()) -> false()
  , not(false()) -> true()
  , odd(0()) -> false()
  , odd(s(x)) -> not(odd(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y))
  , +(s(x), y) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping

 safe(not) = {1}, safe(true) = {}, safe(false) = {}, safe(odd) = {},
 safe(0) = {}, safe(s) = {1}, safe(+) = {}

and precedence

 odd > not .

Following symbols are considered recursive:

 {odd, +}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

   not(; true()) > false()       
                                 
  not(; false()) > true()        
                                 
       odd(0();) > false()       
                                 
    odd(s(; x);) > not(; odd(x;))
                                 
     +(x,  0();) > x             
                                 
  +(x,  s(; y);) > s(; +(x,  y;))
                                 
  +(s(; x),  y;) > s(; +(x,  y;))
                                 

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