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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          append#_A(x1,x2) = max{5, x1 + 4}
          ifappend#_A(x1,x2,x3) = max{x1 + 4, x3}
          is_empty_A(x1) = max{5, x1 + 4}
          false_A = 6
          tl_A(x1) = max{0, x1 - 1}
          nil_A = 1
          true_A = 0
          cons_A(x1,x2) = max{x1 + 7, x2 + 7}
    
      precedence:
      
        append# > ifappend# > nil > cons > true > is_empty = false = tl
      
      partial status:
    
        pi(append#) = [1]
        pi(ifappend#) = [1, 3]
        pi(is_empty) = [1]
        pi(false) = []
        pi(tl) = []
        pi(nil) = []
        pi(true) = []
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        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 + 3
          false_A = 5
          tl_A(x1) = 1
          nil_A = 3
          true_A = 4
          cons_A(x1,x2) = max{x1 + 6, x2 + 6}
    
      precedence:
      
        append# > is_empty = false = tl = true > nil = cons > ifappend#
      
      partial status:
    
        pi(append#) = [1]
        pi(ifappend#) = []
        pi(is_empty) = []
        pi(false) = []
        pi(tl) = []
        pi(nil) = []
        pi(true) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1, p2

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