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

Strict Trs:
  { plus_x#1(0(), x8) -> x8
  , plus_x#1(S(x12), x14) -> S(plus_x#1(x12, x14))
  , map#2(plus_x(x2), Nil()) -> Nil()
  , map#2(plus_x(x6), Cons(x4, x2)) ->
    Cons(plus_x#1(x6, x4), map#2(plus_x(x6), x2))
  , main(x5, x12) -> map#2(plus_x(x12), x5) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add the following weak dependency pairs:

Strict DPs:
  { plus_x#1^#(0(), x8) -> c_1()
  , plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))
  , main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }

and mark the set of starting terms.

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

Strict DPs:
  { plus_x#1^#(0(), x8) -> c_1()
  , plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))
  , main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }
Strict Trs:
  { plus_x#1(0(), x8) -> x8
  , plus_x#1(S(x12), x14) -> S(plus_x#1(x12, x14))
  , map#2(plus_x(x2), Nil()) -> Nil()
  , map#2(plus_x(x6), Cons(x4, x2)) ->
    Cons(plus_x#1(x6, x4), map#2(plus_x(x6), x2))
  , main(x5, x12) -> map#2(plus_x(x12), x5) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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^2)).

Strict DPs:
  { plus_x#1^#(0(), x8) -> c_1()
  , plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))
  , main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

                   [0] = [0]                      
                         [0]                      
                                                  
               [S](x1) = [1 2] x1 + [0]           
                         [0 1]      [0]           
                                                  
          [plus_x](x1) = [1 0] x1 + [0]           
                         [0 0]      [0]           
                                                  
                 [Nil] = [0]                      
                         [0]                      
                                                  
        [Cons](x1, x2) = [1 0] x2 + [0]           
                         [0 0]      [0]           
                                                  
  [plus_x#1^#](x1, x2) = [0 0] x1 + [0]           
                         [1 0]      [0]           
                                                  
                 [c_1] = [0]                      
                         [0]                      
                                                  
             [c_2](x1) = [1 0] x1 + [0]           
                         [0 1]      [0]           
                                                  
     [map#2^#](x1, x2) = [0]                      
                         [0]                      
                                                  
                 [c_3] = [0]                      
                         [0]                      
                                                  
         [c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                         [0 1]      [0 1]      [0]
                                                  
      [main^#](x1, x2) = [1 2] x1 + [2 2] x2 + [2]
                         [1 2]      [2 2]      [2]
                                                  
             [c_5](x1) = [1 0] x1 + [0]           
                         [0 1]      [2]           

The order satisfies the following ordering constraints:

                [plus_x#1^#(0(), x8)] =  [0]                                               
                                         [0]                                               
                                      >= [0]                                               
                                         [0]                                               
                                      =  [c_1()]                                           
                                                                                           
            [plus_x#1^#(S(x12), x14)] =  [0 0] x12 + [0]                                   
                                         [1 2]       [0]                                   
                                      >= [0 0] x12 + [0]                                   
                                         [1 0]       [0]                                   
                                      =  [c_2(plus_x#1^#(x12, x14))]                       
                                                                                           
         [map#2^#(plus_x(x2), Nil())] =  [0]                                               
                                         [0]                                               
                                      >= [0]                                               
                                         [0]                                               
                                      =  [c_3()]                                           
                                                                                           
  [map#2^#(plus_x(x6), Cons(x4, x2))] =  [0]                                               
                                         [0]                                               
                                      ?  [0 0] x6 + [0]                                    
                                         [1 0]      [0]                                    
                                      =  [c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))]
                                                                                           
                    [main^#(x5, x12)] =  [2 2] x12 + [1 2] x5 + [2]                        
                                         [2 2]       [1 2]      [2]                        
                                      >  [0]                                               
                                         [2]                                               
                                      =  [c_5(map#2^#(plus_x(x12), x5))]                   
                                                                                           

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^2)).

Strict DPs:
  { plus_x#1^#(0(), x8) -> c_1()
  , plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
Weak DPs: { main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We estimate the number of application of {1} by applications of
Pre({1}) = {2,4}. Here rules are labeled as follows:

  DPs:
    { 1: plus_x#1^#(0(), x8) -> c_1()
    , 2: plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
    , 3: map#2^#(plus_x(x2), Nil()) -> c_3()
    , 4: map#2^#(plus_x(x6), Cons(x4, x2)) ->
         c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))
    , 5: main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }

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

Strict DPs:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
Weak DPs:
  { plus_x#1^#(0(), x8) -> c_1()
  , main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

{ plus_x#1^#(0(), x8) -> c_1() }

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

Strict DPs:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
Weak DPs: { main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

Consider the dependency graph

  1: plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
     -->_1 plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14)) :1
  
  2: map#2^#(plus_x(x2), Nil()) -> c_3()
  
  3: map#2^#(plus_x(x6), Cons(x4, x2)) ->
     c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2))
     -->_2 map#2^#(plus_x(x6), Cons(x4, x2)) ->
           c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) :3
     -->_2 map#2^#(plus_x(x2), Nil()) -> c_3() :2
     -->_1 plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14)) :1
  
  4: main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5))
     -->_1 map#2^#(plus_x(x6), Cons(x4, x2)) ->
           c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) :3
     -->_1 map#2^#(plus_x(x2), Nil()) -> c_3() :2
  

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).

  { main^#(x5, x12) -> c_5(map#2^#(plus_x(x12), x5)) }


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

Strict DPs:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x2), Nil()) -> c_3()
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We estimate the number of application of {2} by applications of
Pre({2}) = {3}. Here rules are labeled as follows:

  DPs:
    { 1: plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
    , 2: map#2^#(plus_x(x2), Nil()) -> c_3()
    , 3: map#2^#(plus_x(x6), Cons(x4, x2)) ->
         c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }

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

Strict DPs:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
Weak DPs: { map#2^#(plus_x(x2), Nil()) -> c_3() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

{ map#2^#(plus_x(x2), Nil()) -> c_3() }

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

Strict DPs:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
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.

DPs:
  { 1: plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , 2: map#2^#(plus_x(x6), Cons(x4, x2)) ->
       c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,2-bounded)' as induced by the safe mapping
  
   safe(S) = {1}, safe(plus_x) = {1}, safe(Cons) = {1, 2},
   safe(plus_x#1^#) = {2}, safe(c_2) = {}, safe(map#2^#) = {},
   safe(c_4) = {}
  
  and precedence
  
   map#2^# > plus_x#1^# .
  
  Following symbols are considered recursive:
  
   {plus_x#1^#, map#2^#}
  
  The recursion depth is 2.
  
  Further, following argument filtering is employed:
  
   pi(S) = [1], pi(plus_x) = 1, pi(Cons) = [2], pi(plus_x#1^#) = [1],
   pi(c_2) = [1], pi(map#2^#) = [1, 2], pi(c_4) = [1, 2]
  
  Usable defined function symbols are a subset of:
  
   {plus_x#1^#, map#2^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
              pi(plus_x#1^#(S(x12), x14)) = plus_x#1^#(S(; x12);)                               
                                          > c_2(plus_x#1^#(x12;);)                              
                                          = pi(c_2(plus_x#1^#(x12, x14)))                       
                                                                                                
    pi(map#2^#(plus_x(x6), Cons(x4, x2))) = map#2^#(x6,  Cons(; x2);)                           
                                          > c_4(plus_x#1^#(x6;),  map#2^#(x6,  x2;);)           
                                          = pi(c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)))
                                                                                                

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:
  { plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
  , map#2^#(plus_x(x6), Cons(x4, x2)) ->
    c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }
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.

{ plus_x#1^#(S(x12), x14) -> c_2(plus_x#1^#(x12, x14))
, map#2^#(plus_x(x6), Cons(x4, x2)) ->
  c_4(plus_x#1^#(x6, x4), map#2^#(plus_x(x6), x2)) }

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^2))