YES

We show the termination of the TRS R:

  app(id(),x) -> x
  app(add(),|0|()) -> id()
  app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

-- SCC decomposition.

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

p1: app#(app(add(),app(s(),x)),y) -> app#(s(),app(app(add(),x),y))
p2: app#(app(add(),app(s(),x)),y) -> app#(app(add(),x),y)
p3: app#(app(add(),app(s(),x)),y) -> app#(add(),x)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p5: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p6: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p7: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(id(),x) -> x
r2: app(add(),|0|()) -> id()
r3: app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p2, p6, p7}


-- Reduction pair.

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

p1: app#(app(add(),app(s(),x)),y) -> app#(app(add(),x),y)
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)

and R consists of:

r1: app(id(),x) -> x
r2: app(add(),|0|()) -> id()
r3: app(app(add(),app(s(),x)),y) -> app(s(),app(app(add(),x),y))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The set of usable rules consists of

  r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = x1 + x2
      app_A(x1,x2) = x1 + x2 + (0,1)
      add_A() = (1,1)
      s_A() = (1,1)
      map_A() = (1,1)
      cons_A() = (1,0)
      |0|_A() = (1,1)
      id_A() = (0,3)

The next rules are strictly ordered:

  p1, p2, p3

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