We consider the following Problem:

  Strict Trs:
    {  is_empty(nil()) -> true()
     , is_empty(cons(x, l)) -> false()
     , hd(cons(x, l)) -> x
     , tl(cons(x, l)) -> l
     , append(l1, l2) -> ifappend(l1, l2, l1)
     , ifappend(l1, l2, nil()) -> l2
     , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  is_empty(nil()) -> true()
       , is_empty(cons(x, l)) -> false()
       , hd(cons(x, l)) -> x
       , tl(cons(x, l)) -> l
       , append(l1, l2) -> ifappend(l1, l2, l1)
       , ifappend(l1, l2, nil()) -> l2
       , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    We consider the following Problem:
    
      Strict Trs:
        {  is_empty(nil()) -> true()
         , is_empty(cons(x, l)) -> false()
         , hd(cons(x, l)) -> x
         , tl(cons(x, l)) -> l
         , append(l1, l2) -> ifappend(l1, l2, l1)
         , ifappend(l1, l2, nil()) -> l2
         , ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The following argument positions are usable:
        Uargs(is_empty) = {}, Uargs(cons) = {2}, Uargs(hd) = {},
        Uargs(tl) = {}, Uargs(append) = {}, Uargs(ifappend) = {}
      We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       is_empty(x1) = [0 0] x1 + [1]
                      [0 0]      [1]
       nil() = [0]
               [2]
       true() = [0]
                [0]
       cons(x1, x2) = [0 1] x1 + [1 0] x2 + [2]
                      [1 0]      [0 1]      [0]
       false() = [0]
                 [0]
       hd(x1) = [0 1] x1 + [1]
                [1 0]      [3]
       tl(x1) = [1 0] x1 + [3]
                [1 1]      [3]
       append(x1, x2) = [2 1] x1 + [2 0] x2 + [1]
                        [1 1]      [0 2]      [2]
       ifappend(x1, x2, x3) = [0 0] x1 + [2 0] x2 + [2 1] x3 + [0]
                              [0 0]      [0 2]      [1 1]      [1]

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