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

  (no rules)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x1 + 5
              app#_A(x1,x2) = x1
              map_1_A(x1) = x1 + 3
              map_2#!_A(x1,x2) = x1 + 2
              cons_2!_A(x1,x2) = 1
              app#!_A(x1,x2) = x1
              map_1!_A(x1) = x1 + 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x1 + x2
              map_1_A(x1) = x1 + 3
              map_2#!_A(x1,x2) = 6
              cons_2!_A(x1,x2) = 1
              app#!_A(x1,x2) = x1 + 2
              map_1!_A(x1) = 7
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]

The following rules are removed:

  p1, p2



-- SCC decomposition.

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 estimated dependency graph contains the following SCCs:

  {p1}


-- 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, >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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 2
              map_2#!_A(x1,x2) = 6
              cons_2!_A(x1,x2) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 2
              map_2#!_A(x1,x2) = x2 + 5
              cons_2!_A(x1,x2) = 4
        
  
    partial status:
  
      pi(map_2#) = [1]
      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

  (no rules)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              add_2#_A(x1,x2) = x1
              s_1_A(x1) = x1 + 2
              add_2#!_A(x1,x2) = x1 + 2
              s_1!_A(x1) = x1 + 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              add_2#_A(x1,x2) = 0
              s_1_A(x1) = 3
              add_2#!_A(x1,x2) = 0
              s_1!_A(x1) = 1
        
  
    partial status:
  
      pi(add_2#) = []
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.