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

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      append#_A(x1,x2) = x1 + (1,0)
      ifappend#_A(x1,x2,x3) = x3 + (0,1)
      cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,1)

The next rules are strictly ordered:

  p1, p2

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