YES
We show the termination of the TRS R:

  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))
  app(app(minus(),x),|0|()) -> x
  app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),app(p(),app(s(),x))),app(p(),app(s(),y)))
  app(p(),app(s(),x)) -> x
  app(app(div(),|0|()),app(s(),y)) -> |0|()
  app(app(div(),app(s(),x)),app(s(),y)) -> app(s(),app(app(div(),app(app(minus(),x),app(id(),y))),app(s(),y)))
  app(id(),x) -> x
  app(id(),x) -> app(s(),app(s(),app(s(),x)))
  app(id(),app(p(),x)) -> app(id(),app(s(),app(id(),x)))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(div_0(),y) -> div_1(y)
  app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
  app(minus_1(x1),y) -> minus_2(x1,y)
  app(p_0(),y) -> p_1(y)
  app(s_0(),y) -> s_1(y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  minus_2(x,|0|_0()) -> x
  minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
  p_1(s_1(x)) -> x
  div_2(|0|_0(),s_1(y)) -> |0|_0()
  div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
  id_1(x) -> x
  id_1(x) -> s_1(s_1(s_1(x)))
  id_1(p_1(x)) -> id_1(s_1(id_1(x)))


-- SCC decomposition.

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

p1: app#(div_1(x1),y) -> div_2#(x1,y)
p2: app#(id_0(),y) -> id_1#(y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(minus_1(x1),y) -> minus_2#(x1,y)
p5: app#(p_0(),y) -> p_1#(y)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p8: minus_2#(s_1(x),s_1(y)) -> minus_2#(p_1(s_1(x)),p_1(s_1(y)))
p9: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(x))
p10: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(y))
p11: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,id_1(y)),s_1(y))
p12: div_2#(s_1(x),s_1(y)) -> minus_2#(x,id_1(y))
p13: div_2#(s_1(x),s_1(y)) -> id_1#(y)
p14: id_1#(p_1(x)) -> id_1#(s_1(id_1(x)))
p15: id_1#(p_1(x)) -> id_1#(x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
r9: app(minus_1(x1),y) -> minus_2(x1,y)
r10: app(p_0(),y) -> p_1(y)
r11: app(s_0(),y) -> s_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: minus_2(x,|0|_0()) -> x
r15: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r16: p_1(s_1(x)) -> x
r17: div_2(|0|_0(),s_1(y)) -> |0|_0()
r18: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
r19: id_1(x) -> x
r20: id_1(x) -> s_1(s_1(s_1(x)))
r21: id_1(p_1(x)) -> id_1(s_1(id_1(x)))

The estimated dependency graph contains the following SCCs:

  {p3, p6, p7}
  {p11}
  {p15}
  {p8}


-- 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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
r9: app(minus_1(x1),y) -> minus_2(x1,y)
r10: app(p_0(),y) -> p_1(y)
r11: app(s_0(),y) -> s_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: minus_2(x,|0|_0()) -> x
r15: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r16: p_1(s_1(x)) -> x
r17: div_2(|0|_0(),s_1(y)) -> |0|_0()
r18: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
r19: id_1(x) -> x
r20: id_1(x) -> s_1(s_1(s_1(x)))
r21: id_1(p_1(x)) -> id_1(s_1(id_1(x)))

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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x1 - 6, x2 - 6}
            cons_2_A(x1,x2) = max{5, x1 + 4, x2}
            app#_A(x1,x2) = max{0, x1 - 8, x2 - 2}
            map_1_A(x1) = max{7, x1 + 2}
            map_2#!_A(x1,x2) = max{3, x1 - 2, x2 - 6}
            cons_2!_A(x1,x2) = max{6, x1 + 2, x2 + 4}
            app#!_A(x1,x2) = max{3, x1 - 3, x2 - 5}
            map_1!_A(x1) = max{2, x1 - 76}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x1 - 3, x2 - 35}
            cons_2_A(x1,x2) = max{x1 + 4, x2 + 34}
            app#_A(x1,x2) = max{0, x1 - 3, x2 - 31}
            map_1_A(x1) = x1 + 2
            map_2#!_A(x1,x2) = max{4, x1 - 1, x2 - 35}
            cons_2!_A(x1,x2) = max{3, x1 - 35, x2 - 66}
            app#!_A(x1,x2) = max{4, x1 - 3, x2 - 35}
            map_1!_A(x1) = max{5, x1 - 2}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,id_1(y)),s_1(y))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
r9: app(minus_1(x1),y) -> minus_2(x1,y)
r10: app(p_0(),y) -> p_1(y)
r11: app(s_0(),y) -> s_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: minus_2(x,|0|_0()) -> x
r15: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r16: p_1(s_1(x)) -> x
r17: div_2(|0|_0(),s_1(y)) -> |0|_0()
r18: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
r19: id_1(x) -> x
r20: id_1(x) -> s_1(s_1(s_1(x)))
r21: id_1(p_1(x)) -> id_1(s_1(id_1(x)))

The set of usable rules consists of

  r14, r15, r16

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!):
        
          max/plus interpretation on natural numbers:
          
            div_2#_A(x1,x2) = max{0, x1 - 11, x2 - 10}
            s_1_A(x1) = x1 + 6
            minus_2_A(x1,x2) = max{8, x1}
            id_1_A(x1) = 0
            |0|_0_A = 8
            p_1_A(x1) = max{8, x1 - 6}
            div_2#!_A(x1,x2) = max{x1 + 2, x2 + 5}
            s_1!_A(x1) = max{9, x1 + 6}
            minus_2!_A(x1,x2) = 0
            id_1!_A(x1) = max{13, x1}
            |0|_0!_A = 7
            p_1!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div_2#_A(x1,x2) = 0
            s_1_A(x1) = max{14, x1 + 8}
            minus_2_A(x1,x2) = x1 + 9
            id_1_A(x1) = max{29, x1 - 9}
            |0|_0_A = 13
            p_1_A(x1) = max{11, x1 - 5}
            div_2#!_A(x1,x2) = 2
            s_1!_A(x1) = max{39, x1 + 8}
            minus_2!_A(x1,x2) = max{3, x1 - 4}
            id_1!_A(x1) = max{38, x1 - 38}
            |0|_0!_A = 9
            p_1!_A(x1) = 2
        
  
    partial status:
  
      pi(div_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(id_1) = [1]
      pi(|0|_0) = []
      pi(p_1) = [1]

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: id_1#(p_1(x)) -> id_1#(x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
r9: app(minus_1(x1),y) -> minus_2(x1,y)
r10: app(p_0(),y) -> p_1(y)
r11: app(s_0(),y) -> s_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: minus_2(x,|0|_0()) -> x
r15: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r16: p_1(s_1(x)) -> x
r17: div_2(|0|_0(),s_1(y)) -> |0|_0()
r18: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
r19: id_1(x) -> x
r20: id_1(x) -> s_1(s_1(s_1(x)))
r21: id_1(p_1(x)) -> id_1(s_1(id_1(x)))

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!):
        
          max/plus interpretation on natural numbers:
          
            id_1#_A(x1) = x1
            p_1_A(x1) = max{8, x1 + 3}
            id_1#!_A(x1) = max{4, x1 - 81}
            p_1!_A(x1) = max{4, x1 - 80}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            id_1#_A(x1) = x1
            p_1_A(x1) = max{5, x1 + 1}
            id_1#!_A(x1) = max{3, x1 - 63}
            p_1!_A(x1) = max{3, x1 - 62}
        
  
    partial status:
  
      pi(id_1#) = [1]
      pi(p_1) = [1]

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: minus_2#(s_1(x),s_1(y)) -> minus_2#(p_1(s_1(x)),p_1(s_1(y)))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_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(minus_0(),y) -> minus_1(y)
r9: app(minus_1(x1),y) -> minus_2(x1,y)
r10: app(p_0(),y) -> p_1(y)
r11: app(s_0(),y) -> s_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: minus_2(x,|0|_0()) -> x
r15: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r16: p_1(s_1(x)) -> x
r17: div_2(|0|_0(),s_1(y)) -> |0|_0()
r18: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,id_1(y)),s_1(y)))
r19: id_1(x) -> x
r20: id_1(x) -> s_1(s_1(s_1(x)))
r21: id_1(p_1(x)) -> id_1(s_1(id_1(x)))

The set of usable rules consists of

  r16

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!):
        
          max/plus interpretation on natural numbers:
          
            minus_2#_A(x1,x2) = max{0, x1 - 25, x2 - 10}
            s_1_A(x1) = x1 + 9
            p_1_A(x1) = max{8, x1 - 5}
            minus_2#!_A(x1,x2) = x2 + 4
            s_1!_A(x1) = 3
            p_1!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus_2#_A(x1,x2) = max{0, x1 - 19, x2 - 13}
            s_1_A(x1) = x1 + 5
            p_1_A(x1) = max{12, x1}
            minus_2#!_A(x1,x2) = 0
            s_1!_A(x1) = 4
            p_1!_A(x1) = max{0, x1 - 26}
        
  
    partial status:
  
      pi(minus_2#) = [1]
      pi(s_1) = [1]
      pi(p_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.