YES

We show the termination of the TRS R:

  app(app(append(),nil()),l) -> l
  app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t))
  app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3))
  app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))

-- SCC decomposition.

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

p1: app#(app(append(),app(app(cons(),h),t)),l) -> app#(app(cons(),h),app(app(append(),t),l))
p2: app#(app(append(),app(app(cons(),h),t)),l) -> app#(app(append(),t),l)
p3: app#(app(append(),app(app(cons(),h),t)),l) -> app#(append(),t)
p4: app#(app(map(),f),app(app(cons(),h),t)) -> app#(app(cons(),app(f,h)),app(app(map(),f),t))
p5: app#(app(map(),f),app(app(cons(),h),t)) -> app#(cons(),app(f,h))
p6: app#(app(map(),f),app(app(cons(),h),t)) -> app#(f,h)
p7: app#(app(map(),f),app(app(cons(),h),t)) -> app#(app(map(),f),t)
p8: app#(app(append(),app(app(append(),l1),l2)),l3) -> app#(app(append(),l1),app(app(append(),l2),l3))
p9: app#(app(append(),app(app(append(),l1),l2)),l3) -> app#(app(append(),l2),l3)
p10: app#(app(append(),app(app(append(),l1),l2)),l3) -> app#(append(),l2)
p11: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))
p12: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(append(),app(app(map(),f),l1))
p13: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(app(map(),f),l1)
p14: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(app(map(),f),l2)

and R consists of:

r1: app(app(append(),nil()),l) -> l
r2: app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l))
r3: app(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t))
r5: app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3))
r6: app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))

The estimated dependency graph contains the following SCCs:

  {p6, p7, p13, p14}
  {p2, p8, p9}


-- Reduction pair.

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

p1: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(app(map(),f),l2)
p2: app#(app(map(),f),app(app(append(),l1),l2)) -> app#(app(map(),f),l1)
p3: app#(app(map(),f),app(app(cons(),h),t)) -> app#(app(map(),f),t)
p4: app#(app(map(),f),app(app(cons(),h),t)) -> app#(f,h)

and R consists of:

r1: app(app(append(),nil()),l) -> l
r2: app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l))
r3: app(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t))
r5: app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3))
r6: app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > app# > map > cons > append
    
    argument filter:
  
      pi(app#) = [1, 2]
      pi(app) = [1, 2]
      pi(map) = []
      pi(append) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1, p2, p3, p4
  r1, r2, r3, r4, r5, r6

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

-- Reduction pair.

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

p1: app#(app(append(),app(app(cons(),h),t)),l) -> app#(app(append(),t),l)
p2: app#(app(append(),app(app(append(),l1),l2)),l3) -> app#(app(append(),l2),l3)
p3: app#(app(append(),app(app(append(),l1),l2)),l3) -> app#(app(append(),l1),app(app(append(),l2),l3))

and R consists of:

r1: app(app(append(),nil()),l) -> l
r2: app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l))
r3: app(app(map(),f),nil()) -> nil()
r4: app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t))
r5: app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3))
r6: app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))

The set of usable rules consists of

  r1, r2, r5

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      nil > app > append > app# > cons
    
    argument filter:
  
      pi(app#) = 1
      pi(app) = [1, 2]
      pi(append) = []
      pi(cons) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1, p2, p3

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