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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            append#_A(x1,x2) = max{0, x1 - 2, x2 - 8}
            ifappend#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 8, x3 - 7}
            is_empty_A(x1) = x1 + 2
            false_A = 5
            tl_A(x1) = max{1, x1 - 1}
            nil_A = 0
            true_A = 0
            cons_A(x1,x2) = max{x1 - 3, x2 + 6}
            append#!_A(x1,x2) = x1 + 3
            ifappend#!_A(x1,x2,x3) = max{x1 + 2, x3 - 1}
            is_empty!_A(x1) = x1 + 9
            false!_A = 5
            tl!_A(x1) = max{8, x1 - 16}
            nil!_A = 6
            true!_A = 5
            cons!_A(x1,x2) = max{6, x1 - 3, x2 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            append#_A(x1,x2) = max{0, x1 - 6, x2 - 3}
            ifappend#_A(x1,x2,x3) = max{0, x1 - 6, x2 - 3, x3 - 2}
            is_empty_A(x1) = max{0, x1 - 4}
            false_A = 0
            tl_A(x1) = max{5, x1 - 1}
            nil_A = 0
            true_A = 0
            cons_A(x1,x2) = max{7, x1 - 1, x2 + 3}
            append#!_A(x1,x2) = max{0, x1 - 5}
            ifappend#!_A(x1,x2,x3) = max{0, x1 - 5, x3 - 6}
            is_empty!_A(x1) = max{0, x1 - 6}
            false!_A = 0
            tl!_A(x1) = max{2, x1 - 6}
            nil!_A = 1
            true!_A = 0
            cons!_A(x1,x2) = max{2, x1 - 9, x2 - 3}
        
  
    partial status:
  
      pi(append#) = []
      pi(ifappend#) = []
      pi(is_empty) = [1]
      pi(false) = []
      pi(tl) = [1]
      pi(nil) = []
      pi(true) = []
      pi(cons) = [1, 2]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)