YES(O(1),O(1))

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

Strict Trs:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

We add the following innermost weak dependency pairs:

Strict DPs:
  { first^#(0(), X) -> c_1()
  , first^#(s(X), cons(Y)) -> c_2()
  , from^#(X) -> c_3() }

and mark the set of starting terms.

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

Strict DPs:
  { first^#(0(), X) -> c_1()
  , first^#(s(X), cons(Y)) -> c_2()
  , from^#(X) -> c_3() }
Strict Trs:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(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(1)).

Strict DPs:
  { first^#(0(), X) -> c_1()
  , first^#(s(X), cons(Y)) -> c_2()
  , from^#(X) -> c_3() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

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

The following argument positions are usable:
  none

TcT has computed the following constructor-restricted matrix
interpretation.

                [0] = [2]                      
                      [1]                      
                                               
            [s](x1) = [1]                      
                      [1]                      
                                               
         [cons](x1) = [1]                      
                      [2]                      
                                               
  [first^#](x1, x2) = [1 1] x1 + [1 1] x2 + [1]
                      [1 1]      [1 1]      [1]
                                               
              [c_1] = [1]                      
                      [0]                      
                                               
              [c_2] = [1]                      
                      [1]                      
                                               
       [from^#](x1) = [2]                      
                      [1]                      
                                               
              [c_3] = [1]                      
                      [1]                      

The following symbols are considered usable

  {first^#, from^#}

The order satisfies the following ordering constraints:

         [first^#(0(), X)] = [1 1] X + [4]
                             [1 1]     [4]
                           > [1]          
                             [0]          
                           = [c_1()]      
                                          
  [first^#(s(X), cons(Y))] = [6]          
                             [6]          
                           > [1]          
                             [1]          
                           = [c_2()]      
                                          
               [from^#(X)] = [2]          
                             [1]          
                           > [1]          
                             [1]          
                           = [c_3()]      
                                          

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(1)).

Weak DPs:
  { first^#(0(), X) -> c_1()
  , first^#(s(X), cons(Y)) -> c_2()
  , from^#(X) -> c_3() }
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.

{ first^#(0(), X) -> c_1()
, first^#(s(X), cons(Y)) -> c_2()
, from^#(X) -> c_3() }

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(1))