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

Strict Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k))
  , g(a, b, c) -> f(a, cons(b, c)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following weak dependency pairs:

Strict DPs:
  { f^#(empty(), l) -> c_1()
  , f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
  , g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }

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:
  { f^#(empty(), l) -> c_1()
  , f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
  , g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
Strict Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k))
  , g(a, b, c) -> f(a, cons(b, c)) }
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:
  { f^#(empty(), l) -> c_1()
  , f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
  , g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
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_2) = {1}, Uargs(c_3) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

            [empty] = [0]           
                      [0]           
                                    
     [cons](x1, x2) = [0]           
                      [0]           
                                    
      [f^#](x1, x2) = [0]           
                      [0]           
                                    
              [c_1] = [0]           
                      [0]           
                                    
          [c_2](x1) = [1 0] x1 + [0]
                      [0 1]      [2]
                                    
  [g^#](x1, x2, x3) = [1]           
                      [0]           
                                    
          [c_3](x1) = [1 0] x1 + [0]
                      [0 1]      [0]

The order satisfies the following ordering constraints:

     [f^#(empty(), l)] =  [0]                         
                          [0]                         
                       >= [0]                         
                          [0]                         
                       =  [c_1()]                     
                                                      
  [f^#(cons(x, k), l)] =  [0]                         
                          [0]                         
                       ?  [1]                         
                          [2]                         
                       =  [c_2(g^#(k, l, cons(x, k)))]
                                                      
        [g^#(a, b, c)] =  [1]                         
                          [0]                         
                       >  [0]                         
                          [0]                         
                       =  [c_3(f^#(a, cons(b, c)))]   
                                                      

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:
  { f^#(empty(), l) -> c_1()
  , f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k))) }
Weak DPs: { g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
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: f^#(empty(), l) -> c_1() }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_2) = {1}, Uargs(c_3) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
              [empty] = [1]         
                                    
       [cons](x1, x2) = [1] x2 + [0]
                                    
        [f^#](x1, x2) = [1] x1 + [0]
                                    
                [c_1] = [0]         
                                    
            [c_2](x1) = [1] x1 + [0]
                                    
    [g^#](x1, x2, x3) = [1] x1 + [0]
                                    
            [c_3](x1) = [1] x1 + [0]
  
  The order satisfies the following ordering constraints:
  
       [f^#(empty(), l)] =  [1]                         
                         >  [0]                         
                         =  [c_1()]                     
                                                        
    [f^#(cons(x, k), l)] =  [1] k + [0]                 
                         >= [1] k + [0]                 
                         =  [c_2(g^#(k, l, cons(x, k)))]
                                                        
          [g^#(a, b, c)] =  [1] a + [0]                 
                         >= [1] a + [0]                 
                         =  [c_3(f^#(a, cons(b, c)))]   
                                                        

The strictly oriented rules are moved into the weak component.

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

Strict DPs: { f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k))) }
Weak DPs:
  { f^#(empty(), l) -> c_1()
  , g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
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.

{ f^#(empty(), l) -> c_1() }

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

Strict DPs: { f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k))) }
Weak DPs: { g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
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: f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
  , 2: g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_2) = {1}, Uargs(c_3) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
              [empty] = [0]         
                                    
       [cons](x1, x2) = [1] x2 + [1]
                                    
        [f^#](x1, x2) = [4] x1 + [0]
                                    
                [c_1] = [0]         
                                    
            [c_2](x1) = [1] x1 + [0]
                                    
    [g^#](x1, x2, x3) = [4] x1 + [1]
                                    
            [c_3](x1) = [1] x1 + [0]
  
  The order satisfies the following ordering constraints:
  
    [f^#(cons(x, k), l)] = [4] k + [4]                 
                         > [4] k + [1]                 
                         = [c_2(g^#(k, l, cons(x, k)))]
                                                       
          [g^#(a, b, c)] = [4] a + [1]                 
                         > [4] a + [0]                 
                         = [c_3(f^#(a, cons(b, c)))]   
                                                       

The strictly oriented rules are moved into the weak component.

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

Weak DPs:
  { f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
  , g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }
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.

{ f^#(cons(x, k), l) -> c_2(g^#(k, l, cons(x, k)))
, g^#(a, b, c) -> c_3(f^#(a, cons(b, c))) }

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))