YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { exp(x, 0()) -> s(0())
  , exp(x, s(y)) -> *(x, exp(x, y))
  , *(0(), y) -> 0()
  , *(s(x), y) -> +(y, *(x, y))
  , -(x, 0()) -> x
  , -(0(), y) -> 0()
  , -(s(x), s(y)) -> -(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'Small Polynomial Path Order (PS,2-bounded)'
to orient following rules strictly.

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

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . 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,2-bounded)' as induced by the safe mapping
  
   safe(exp) = {}, safe(0) = {}, safe(s) = {1}, safe(*) = {2},
   safe(+) = {1, 2}, safe(-) = {2}
  
  and precedence
  
   exp > *, - > *, exp ~ - .
  
  Following symbols are considered recursive:
  
   {exp, *, -}
  
  The recursion depth is 2.
  
  For your convenience, here are the satisfied ordering constraints:
  
        exp(x,  0();) > s(; 0())         
                                         
     exp(x,  s(; y);) > *(x; exp(x,  y;))
                                         
            *(0(); y) > 0()              
                                         
         *(s(; x); y) > +(; y,  *(x; y)) 
                                         
            -(x; 0()) > x                
                                         
            -(0(); y) > 0()              
                                         
    -(s(; x); s(; y)) > -(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:
  { exp(x, 0()) -> s(0())
  , exp(x, s(y)) -> *(x, exp(x, y))
  , *(0(), y) -> 0()
  , *(s(x), y) -> +(y, *(x, y))
  , -(x, 0()) -> x
  , -(0(), y) -> 0()
  , -(s(x), s(y)) -> -(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^2))