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))

Instead we prove termination of the uncurried system:

  app(add_0(),y) -> add_1(y)
  app(add_1(x1),y) -> add_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(id_0(),y) -> id_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(s_0(),y) -> s_1(y)
  id_1(x) -> x
  add_1(|0|_0()) -> id_0()
  add_2(|0|_0(),y1) -> id_1(y1)
  add_2(s_1(x),y) -> s_1(add_2(x,y))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))


-- SCC decomposition.

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

p1: app#(add_0(),y) -> add_1#(y)
p2: app#(add_1(x1),y) -> add_2#(x1,y)
p3: app#(id_0(),y) -> id_1#(y)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: add_2#(|0|_0(),y1) -> id_1#(y1)
p6: add_2#(s_1(x),y) -> add_2#(x,y)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(add_0(),y) -> add_1(y)
r2: app(add_1(x1),y) -> add_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(id_0(),y) -> id_1(y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: id_1(x) -> x
r10: add_1(|0|_0()) -> id_0()
r11: add_2(|0|_0(),y1) -> id_1(y1)
r12: add_2(s_1(x),y) -> s_1(add_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p4, p7, p8}
  {p6}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(add_0(),y) -> add_1(y)
r2: app(add_1(x1),y) -> add_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(id_0(),y) -> id_1(y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: id_1(x) -> x
r10: add_1(|0|_0()) -> id_0()
r11: add_2(|0|_0(),y1) -> id_1(y1)
r12: add_2(s_1(x),y) -> s_1(add_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The set of usable rules consists of

  r3, r4, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x1 + 5
        cons_2_A(x1,x2) = max{2, x1 - 23, x2 - 23}
        app#_A(x1,x2) = max{4, x1 - 1}
        map_1_A(x1) = max{10, x1 + 7}
        app_A(x1,x2) = max{13, x1 - 8, x2 - 13}
        cons_0_A = 10
        cons_1_A(x1) = max{1, x1 - 14}
        s_0_A = 7
        s_1_A(x1) = max{9, x1 - 24}
  
    base order 2:
  
      precedence:
      
        map_1 > map_2# = app > app# > cons_2 = s_1 > cons_1 = s_0 > cons_0
      
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(s_0) = []
      pi(s_1) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(add_0(),y) -> add_1(y)
r2: app(add_1(x1),y) -> add_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(id_0(),y) -> id_1(y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: id_1(x) -> x
r10: add_1(|0|_0()) -> id_0()
r11: add_2(|0|_0(),y1) -> id_1(y1)
r12: add_2(s_1(x),y) -> s_1(add_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(add_0(),y) -> add_1(y)
r2: app(add_1(x1),y) -> add_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(id_0(),y) -> id_1(y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: id_1(x) -> x
r10: add_1(|0|_0()) -> id_0()
r11: add_2(|0|_0(),y1) -> id_1(y1)
r12: add_2(s_1(x),y) -> s_1(add_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{4, x2 + 3}
        cons_2_A(x1,x2) = max{x1 + 3, x2 + 5}
  
    base order 2:
  
      precedence:
      
        cons_2 > map_2#
      
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: add_2#(s_1(x),y) -> add_2#(x,y)

and R consists of:

r1: app(add_0(),y) -> add_1(y)
r2: app(add_1(x1),y) -> add_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(id_0(),y) -> id_1(y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: id_1(x) -> x
r10: add_1(|0|_0()) -> id_0()
r11: add_2(|0|_0(),y1) -> id_1(y1)
r12: add_2(s_1(x),y) -> s_1(add_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The set of usable rules consists of

  r1, r3, r4, r5, r8, r9, r10, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        add_2#_A(x1,x2) = max{x1 + 4, x2 + 9}
        s_1_A(x1) = max{7, x1 + 5}
        app_A(x1,x2) = max{15, x1 - 18, x2 + 6}
        add_0_A = 14
        add_1_A(x1) = max{15, x1 - 7}
        cons_0_A = 17
        cons_1_A(x1) = max{7, x1 - 1}
        cons_2_A(x1,x2) = max{8, x1 - 20, x2 - 28}
        id_0_A = 5
        id_1_A(x1) = max{7, x1 + 1}
        s_0_A = 6
        |0|_0_A = 6
        map_2_A(x1,x2) = max{15, x1 - 38, x2 + 6}
        nil_0_A = 16
  
    base order 2:
  
      precedence:
      
        add_2# = app > s_1 = add_1 = id_1 = map_2 > add_0 = cons_2 = s_0 = |0|_0 = nil_0 > cons_1 = id_0 > cons_0
      
  
    partial status:
  
      pi(add_2#) = [1]
      pi(s_1) = [1]
      pi(app) = []
      pi(add_0) = []
      pi(add_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(id_0) = []
      pi(id_1) = []
      pi(s_0) = []
      pi(|0|_0) = []
      pi(map_2) = [2]
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.