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 reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(1,0,0),(0,0,0)) x1 + ((1,0,0),(1,1,0),(1,0,1)) x2
      app_A(x1,x2) = x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (0,1,1)
      map_A() = (1,1,1)
      append_A() = (1,1,0)
      cons_A() = (1,1,0)

The next rules are strictly ordered:

  p1, p2, p3, p4

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x1 + x2
      app_A(x1,x2) = x1 + x2 + (0,1,0)
      append_A() = (1,1,1)
      cons_A() = (1,1,0)
      nil_A() = (1,1,0)

The next rules are strictly ordered:

  p1, p2, p3

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