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),y)),app(s(),y)))

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(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,y),s_1(y)))


-- 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#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(minus_1(x1),y) -> minus_2#(x1,y)
p4: app#(p_0(),y) -> p_1#(y)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(p_1(s_1(x)),p_1(s_1(y)))
p8: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(x))
p9: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(y))
p10: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(y))
p11: div_2#(s_1(x),s_1(y)) -> minus_2#(x,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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p6}
  {p10}
  {p7}


-- 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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = 1
        cons_2_A(x1,x2) = max{2, x1 - 35, x2 - 35}
        app#_A(x1,x2) = 1
        map_1_A(x1) = max{2, x1 - 42}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{0, x1 - 1, x2 - 21}
          cons_2_A(x1,x2) = max{x1 + 12, x2 + 20}
          app#_A(x1,x2) = max{0, x1 - 11, x2 - 9}
          map_1_A(x1) = x1 + 10
          map_2#!_A(x1,x2) = max{x1 + 6, x2 - 21}
          cons_2!_A(x1,x2) = max{4, x1 - 19, x2 - 17}
          app#!_A(x1,x2) = max{5, x1 - 3, x2 - 20}
          map_1!_A(x1) = max{2, x1 - 19}
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_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(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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

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(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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x2 + 2
        cons_2_A(x1,x2) = max{5, x1, x2 + 3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{5, x1 - 4, x2 - 46}
          map_2#!_A(x1,x2) = max{52, x2 + 47}
          cons_2!_A(x1,x2) = max{3, x1 - 56, x2 - 101}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [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: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r13, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div_2#_A(x1,x2) = max{32, x1 + 1, x2 - 13}
        s_1_A(x1) = x1 + 22
        minus_2_A(x1,x2) = max{31, x1 + 10, x2 + 7}
        |0|_0_A = 11
        p_1_A(x1) = max{4, x1 - 21}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div_2#_A(x1,x2) = max{0, x1 - 7, x2 - 12}
          s_1_A(x1) = max{11, x1 + 6}
          minus_2_A(x1,x2) = max{8, x1 + 2}
          |0|_0_A = 3
          p_1_A(x1) = max{7, x1 - 5}
          div_2#!_A(x1,x2) = max{11, x1 + 6}
          s_1!_A(x1) = x1 + 21
          minus_2!_A(x1,x2) = max{26, x1 - 12, x2 - 14}
          |0|_0!_A = 1
          p_1!_A(x1) = max{1, x1 - 16}
  
    partial status:
  
      pi(div_2#) = [1]
      pi(s_1) = []
      pi(minus_2) = [1, 2]
      pi(|0|_0) = []
      pi(p_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(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(p_0(),y) -> p_1(y)
r10: app(s_0(),y) -> s_1(y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: minus_2(x,|0|_0()) -> x
r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y)))
r15: p_1(s_1(x)) -> x
r16: div_2(|0|_0(),s_1(y)) -> |0|_0()
r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus_2#_A(x1,x2) = 1
        s_1_A(x1) = max{6, x1 - 15}
        p_1_A(x1) = max{23, x1 + 16}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus_2#_A(x1,x2) = max{0, x1 - 22, x2 - 12}
          s_1_A(x1) = x1 + 11
          p_1_A(x1) = max{2, x1 - 7}
          minus_2#!_A(x1,x2) = max{5, x1 - 13, x2 - 5}
          s_1!_A(x1) = 6
          p_1!_A(x1) = max{8, x1 - 14}
  
    partial status:
  
      pi(minus_2#) = []
      pi(s_1) = []
      pi(p_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.