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:
  runtime complexity
Answer:
  YES(O(1),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(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]
                   [0]
                      
  [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [0]
                                            
     [rev^#](x1) = [1 2] x1 + [2]
                   [1 1]      [2]
                                 
       [c_1](x1) = [1 0] x1 + [1]
                   [0 1]      [2]
                                 
  [r1^#](x1, x2) = [0]
                   [0]
                      
           [c_2] = [1]
                   [0]
                      
       [c_3](x1) = [1 0] x1 + [1]
                   [0 1]      [0]

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(1),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(1),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(1),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(1),O(n^1))

We estimate the number of application of {1} by applications of
Pre({1}) = {2}. Here rules are labeled as follows:

  DPs:
    { 1: r1^#(empty(), a) -> c_2()
    , 2: r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }

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

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

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ r1^#(empty(), a) -> c_2() }

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

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 1: r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_3) = {1}
  
  TcT has computed following constructor-based matrix interpretation
  satisfying not(EDA).
  
         [rev](x1) = [1] x1 + [0]
                                 
           [empty] = [0]
                        
      [r1](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [2]
                                          
       [rev^#](x1) = [1] x1 + [0]
                                 
         [c_1](x1) = [1] x1 + [0]
                                 
    [r1^#](x1, x2) = [2] x1 + [0]
                                 
             [c_2] = [0]
                        
         [c_3](x1) = [1] x1 + [0]
  
  This order satisfies following ordering constraints
  
    [r1^#(cons(x, k), a)] = [2] k + [2] x + [4]       
                          > [2] k + [0]               
                          = [c_3(r1^#(k, cons(x, a)))]
                                                      

The strictly oriented rules are moved into the corresponding weak
component(s).

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

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

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ r1^#(cons(x, k), a) -> c_3(r1^#(k, cons(x, a))) }

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

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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