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:
  { app(nil(), YS) -> YS
  , app(cons(X), YS) -> cons(X)
  , from(X) -> cons(X)
  , zWadr(XS, nil()) -> nil()
  , zWadr(nil(), YS) -> nil()
  , zWadr(cons(X), cons(Y)) -> cons(app(Y, cons(X)))
  , prefix(L) -> cons(nil()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { app^#(nil(), YS) -> c_1()
  , app^#(cons(X), YS) -> c_2()
  , from^#(X) -> c_3()
  , zWadr^#(XS, nil()) -> c_4()
  , zWadr^#(nil(), YS) -> c_5()
  , zWadr^#(cons(X), cons(Y)) -> c_6(app^#(Y, cons(X)))
  , prefix^#(L) -> c_7() }

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:
  { app^#(nil(), YS) -> c_1()
  , app^#(cons(X), YS) -> c_2()
  , from^#(X) -> c_3()
  , zWadr^#(XS, nil()) -> c_4()
  , zWadr^#(nil(), YS) -> c_5()
  , zWadr^#(cons(X), cons(Y)) -> c_6(app^#(Y, cons(X)))
  , prefix^#(L) -> c_7() }
Strict Trs:
  { app(nil(), YS) -> YS
  , app(cons(X), YS) -> cons(X)
  , from(X) -> cons(X)
  , zWadr(XS, nil()) -> nil()
  , zWadr(nil(), YS) -> nil()
  , zWadr(cons(X), cons(Y)) -> cons(app(Y, cons(X)))
  , prefix(L) -> cons(nil()) }
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:
  { app^#(nil(), YS) -> c_1()
  , app^#(cons(X), YS) -> c_2()
  , from^#(X) -> c_3()
  , zWadr^#(XS, nil()) -> c_4()
  , zWadr^#(nil(), YS) -> c_5()
  , zWadr^#(cons(X), cons(Y)) -> c_6(app^#(Y, cons(X)))
  , prefix^#(L) -> c_7() }
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_6) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

              [nil] = [2]                  
                                           
         [cons](x1) = [1] x1 + [2]         
                                           
    [app^#](x1, x2) = [1] x1 + [1] x2 + [2]
                                           
              [c_1] = [1]                  
                                           
              [c_2] = [1]                  
                                           
       [from^#](x1) = [1]                  
                                           
              [c_3] = [0]                  
                                           
  [zWadr^#](x1, x2) = [2] x1 + [1] x2 + [1]
                                           
              [c_4] = [2]                  
                                           
              [c_5] = [0]                  
                                           
          [c_6](x1) = [1] x1 + [2]         
                                           
     [prefix^#](x1) = [2]                  
                                           
              [c_7] = [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(1),O(1)).

Weak DPs:
  { app^#(nil(), YS) -> c_1()
  , app^#(cons(X), YS) -> c_2()
  , from^#(X) -> c_3()
  , zWadr^#(XS, nil()) -> c_4()
  , zWadr^#(nil(), YS) -> c_5()
  , zWadr^#(cons(X), cons(Y)) -> c_6(app^#(Y, cons(X)))
  , prefix^#(L) -> c_7() }
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.

{ app^#(nil(), YS) -> c_1()
, app^#(cons(X), YS) -> c_2()
, from^#(X) -> c_3()
, zWadr^#(XS, nil()) -> c_4()
, zWadr^#(nil(), YS) -> c_5()
, zWadr^#(cons(X), cons(Y)) -> c_6(app^#(Y, cons(X)))
, prefix^#(L) -> c_7() }

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))