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:
  { rev(ls) -> r1(ls, empty())
  , r1(empty(), a) -> a
  , r1(cons(x, k), a) -> r1(k, cons(x, a)) }
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:
  { rev(ls) -> r1(ls, empty())
  , r1(empty(), a) -> a
  , r1(cons(x, k), a) -> r1(k, cons(x, a)) }

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(rev) = {}, safe(r1) = {2}, safe(empty) = {},
   safe(cons) = {1, 2}
  
  and precedence
  
   rev > r1 .
  
  Following symbols are considered recursive:
  
   {r1}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
                rev(ls;) > r1(ls; empty())     
                                               
          r1(empty(); a) > a                   
                                               
    r1(cons(; x,  k); a) > r1(k; cons(; x,  a))
                                               

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:
  { rev(ls) -> r1(ls, empty())
  , r1(empty(), a) -> a
  , r1(cons(x, k), a) -> r1(k, cons(x, a)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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