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 add following weak dependency pairs:

Strict DPs:
  { rev^#(ls) -> c_1(r1^#(ls, empty()))
  , r1^#(empty(), a) -> c_2()
  , r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }

and mark the set of starting terms.

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

Strict DPs:
  { rev^#(ls) -> c_1(r1^#(ls, empty()))
  , r1^#(empty(), a) -> c_2()
  , r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }
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))

No rule is usable, rules are removed from the input problem.

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

Strict DPs:
  { rev^#(ls) -> c_1(r1^#(ls, empty()))
  , r1^#(empty(), a) -> c_2()
  , r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_3) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

         [empty] = [0]         
                               
  [cons](x1, x2) = [1] x2 + [0]
                               
     [rev^#](x1) = [1] x1 + [2]
                               
       [c_1](x1) = [1] x1 + [1]
                               
  [r1^#](x1, x2) = [2] x2 + [0]
                               
           [c_2] = [1]         
                               
       [c_3](x1) = [1] x1 + [1]

This order satisfies following ordering constraints:


Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict DPs:
  { r1^#(empty(), a) -> c_2()
  , r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }
Weak DPs: { rev^#(ls) -> c_1(r1^#(ls, empty())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: r1^#(empty(), a) -> c_2()
  
  2: r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a)))
     -->_1 r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) :2
     -->_1 r1^#(empty(), a) -> c_2() :1
  
  3: rev^#(ls) -> c_1(r1^#(ls, empty()))
     -->_1 r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) :2
     -->_1 r1^#(empty(), a) -> c_2() :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { rev^#(ls) -> c_1(r1^#(ls, empty())) }


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

Strict DPs:
  { r1^#(empty(), a) -> c_2()
  , r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_3) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

         [empty] = [3]                  
                                        
  [cons](x1, x2) = [1] x1 + [1] x2 + [2]
                                        
  [r1^#](x1, x2) = [2] x1 + [0]         
                                        
           [c_2] = [1]                  
                                        
       [c_3](x1) = [1] x1 + [3]         

This order satisfies following ordering constraints:

     [r1^#(empty(), a)] = [6]                       
                        > [1]                       
                        = [c_2()]                   
                                                    
  [r1^#(cons(x, k), a)] = [2] x + [2] k + [4]       
                        > [2] k + [3]               
                        = [c_3(r1^#(k, cons(x, a)))]
                                                    

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