YES
We show the termination of the TRS R:

  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))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: append#(l1,l2) -> ifappend#(l1,l2,l1)
p2: ifappend#(l1,l2,cons(x,l)) -> append#(l,l2)

and R consists of:

r1: is_empty(nil()) -> true()
r2: is_empty(cons(x,l)) -> false()
r3: hd(cons(x,l)) -> x
r4: tl(cons(x,l)) -> l
r5: append(l1,l2) -> ifappend(l1,l2,l1)
r6: ifappend(l1,l2,nil()) -> l2
r7: ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: append#(l1,l2) -> ifappend#(l1,l2,l1)
p2: ifappend#(l1,l2,cons(x,l)) -> append#(l,l2)

and R consists of:

r1: is_empty(nil()) -> true()
r2: is_empty(cons(x,l)) -> false()
r3: hd(cons(x,l)) -> x
r4: tl(cons(x,l)) -> l
r5: append(l1,l2) -> ifappend(l1,l2,l1)
r6: ifappend(l1,l2,nil()) -> l2
r7: ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2))

The set of usable rules consists of

  r3, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        append#_A(x1,x2) = max{5, x2 + 1}
        ifappend#_A(x1,x2,x3) = max{5, x2 + 1}
        cons_A(x1,x2) = max{4, x1 + 2, x2 - 6}
        hd_A(x1) = max{0, x1 - 1}
        tl_A(x1) = x1 + 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          append#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          ifappend#_A(x1,x2,x3) = max{0, x1 - 5, x2 - 1, x3 - 11}
          cons_A(x1,x2) = max{x1 + 10, x2 + 6}
          hd_A(x1) = max{0, x1 - 10}
          tl_A(x1) = max{0, x1 - 6}
          append#!_A(x1,x2) = x1 + 4
          ifappend#!_A(x1,x2,x3) = max{3, x1 + 1, x3 - 1}
          cons!_A(x1,x2) = max{3, x1 - 2, x2 - 7}
          hd!_A(x1) = x1
          tl!_A(x1) = x1
  
    partial status:
  
      pi(append#) = [2]
      pi(ifappend#) = []
      pi(cons) = [1]
      pi(hd) = []
      pi(tl) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.