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,is_empty(l1))
  ifappend(l1,l2,true()) -> l2
  ifappend(l1,l2,false()) -> cons(hd(l1),append(tl(l1),l2))

-- SCC decomposition.

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

p1: append#(l1,l2) -> ifappend#(l1,l2,is_empty(l1))
p2: append#(l1,l2) -> is_empty#(l1)
p3: ifappend#(l1,l2,false()) -> hd#(l1)
p4: ifappend#(l1,l2,false()) -> append#(tl(l1),l2)
p5: ifappend#(l1,l2,false()) -> tl#(l1)

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,is_empty(l1))
r6: ifappend(l1,l2,true()) -> l2
r7: ifappend(l1,l2,false()) -> cons(hd(l1),append(tl(l1),l2))

The estimated dependency graph contains the following SCCs:

  {p1, p4}


-- Reduction pair.

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

p1: append#(l1,l2) -> ifappend#(l1,l2,is_empty(l1))
p2: ifappend#(l1,l2,false()) -> append#(tl(l1),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,is_empty(l1))
r6: ifappend(l1,l2,true()) -> l2
r7: ifappend(l1,l2,false()) -> cons(hd(l1),append(tl(l1),l2))

The set of usable rules consists of

  r1, r2, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      append#_A(x1,x2) = x1 + 2
      ifappend#_A(x1,x2,x3) = max{x1 + 2, x3 + 1}
      is_empty_A(x1) = x1 + 1
      false_A = 3
      tl_A(x1) = max{1, x1 - 1}
      nil_A = 0
      true_A = 0
      cons_A(x1,x2) = max{x1 + 3, x2 + 3}
    
    2. max/plus interpretations on natural numbers:
    
      append#_A(x1,x2) = 1
      ifappend#_A(x1,x2,x3) = 0
      is_empty_A(x1) = x1 + 1
      false_A = 2
      tl_A(x1) = 0
      nil_A = 0
      true_A = 0
      cons_A(x1,x2) = 2
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.