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(),x),y)
  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(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(x,y)
  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: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p6: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p7: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(y))
p8: 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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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, p4, p5}
  {p7}
  {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(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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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) = max{4, x1 + 1}
        cons_2_A(x1,x2) = max{2, x1 - 12, x2 - 12}
        app#_A(x1,x2) = max{3, x1 - 3}
        map_1_A(x1) = x1 + 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{0, x1 - 5, x2 - 4}
          cons_2_A(x1,x2) = max{x1 + 3, x2 + 3}
          app#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          map_1_A(x1) = max{4, x1 + 1}
          map_2#!_A(x1,x2) = max{7, x1 - 6, x2 - 7}
          cons_2!_A(x1,x2) = max{2, x1 - 8, x2 - 6}
          app#!_A(x1,x2) = max{8, x1 - 6, x2 - 6}
          map_1!_A(x1) = max{6, x1 - 6}
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = []
      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(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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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 + 8
        cons_2_A(x1,x2) = max{x1 - 7, x2 + 9}
  
    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{10, x1 - 2, x2 - 3}
          map_2#!_A(x1,x2) = 7
          cons_2!_A(x1,x2) = max{6, x1 - 3, x2 - 17}
  
    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: 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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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

  r12, r13

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{1, x1 - 6}
        s_1_A(x1) = max{27, x1 + 8}
        minus_2_A(x1,x2) = max{11, x1 + 5}
        |0|_0_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div_2#_A(x1,x2) = max{0, x1 - 26, x2 - 12}
          s_1_A(x1) = max{10, x1 + 6}
          minus_2_A(x1,x2) = max{25, x1 + 6, x2 + 6}
          |0|_0_A = 17
          div_2#!_A(x1,x2) = max{6, x2 - 4}
          s_1!_A(x1) = max{5, x1 + 2}
          minus_2!_A(x1,x2) = max{7, x1 - 11, x2 - 11}
          |0|_0!_A = 3
  
    partial status:
  
      pi(div_2#) = []
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(|0|_0) = []

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#(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(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: 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:
      
        minus_2#_A(x1,x2) = max{21, x2}
        s_1_A(x1) = x1 + 21
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus_2#_A(x1,x2) = 0
          s_1_A(x1) = max{3, x1 - 7}
          minus_2#!_A(x1,x2) = x2 + 9
          s_1!_A(x1) = max{13, x1 + 10}
  
    partial status:
  
      pi(minus_2#) = [2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.