YES(?,O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict Trs:
  { times(x, plus(y, s(z))) ->
    plus(times(x, plus(y, times(s(z), 0()))), times(x, s(z)))
  , times(x, s(y)) -> plus(times(x, y), x)
  , times(x, 0()) -> 0()
  , plus(x, s(y)) -> s(plus(x, y))
  , plus(x, 0()) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

Arguments of following rules are not normal-forms:

{ times(x, plus(y, s(z))) ->
  plus(times(x, plus(y, times(s(z), 0()))), times(x, s(z))) }

All above mentioned rules can be savely removed.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict Trs:
  { times(x, s(y)) -> plus(times(x, y), x)
  , times(x, 0()) -> 0()
  , plus(x, s(y)) -> s(plus(x, y))
  , plus(x, 0()) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

 safe(times) = {}, safe(plus) = {1}, safe(s) = {1}, safe(0) = {}

and precedence

 times > plus .

Following symbols are considered recursive:

 {times, plus}

The recursion depth is 2.

For your convenience, here are the satisfied ordering constraints:

  times(x,  s(; y);) > plus(x; times(x,  y;))
                                             
     times(x,  0();) > 0()                   
                                             
     plus(s(; y); x) > s(; plus(y; x))       
                                             
        plus(0(); x) > x                     
                                             

Hurray, we answered YES(?,O(n^2))