YES
We show the termination of the TRS R:

  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
  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(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(inc_0(),y) -> inc_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
  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#(inc_0(),y) -> inc_1#(y)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(plus_1(x1),y) -> plus_2#(x1,y)
p4: plus_2#(s_1(x),y) -> plus_2#(x,y)
p5: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: 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(inc_0(),y) -> inc_1(y)
r4: app(map_0(),y) -> map_1(y)
r5: app(map_1(x1),y) -> map_2(x1,y)
r6: app(plus_0(),y) -> plus_1(y)
r7: app(plus_1(x1),y) -> plus_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: plus_2(|0|_0(),y) -> y
r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
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))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p6, p7}
  {p4}


-- Reduction pair.

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

p1: app#(inc_0(),y) -> inc_1#(y)
p2: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: app#(map_1(x1),y) -> map_2#(x1,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(inc_0(),y) -> inc_1(y)
r4: app(map_0(),y) -> map_1(y)
r5: app(map_1(x1),y) -> map_2(x1,y)
r6: app(plus_0(),y) -> plus_1(y)
r7: app(plus_1(x1),y) -> plus_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: plus_2(|0|_0(),y) -> y
r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
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))

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:
      
        app#_A(x1,x2) = max{38, x1 - 41}
        inc_0_A = 39
        inc_1#_A(x1) = 38
        map_2#_A(x1,x2) = max{38, x1 - 2}
        plus_1_A(x1) = max{1, x1 - 17}
        s_1_A(x1) = max{3, x1 - 17}
        |0|_0_A = 37
        cons_2_A(x1,x2) = max{3, x1 - 8, x2 - 8}
        map_1_A(x1) = x1 + 39
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          inc_0_A = 4
          inc_1#_A(x1) = max{0, x1 - 7}
          map_2#_A(x1,x2) = max{0, x1 - 3, x2 - 16}
          plus_1_A(x1) = max{2, x1 - 1}
          s_1_A(x1) = max{3, x1 - 3}
          |0|_0_A = 1
          cons_2_A(x1,x2) = max{x1 + 15, x2 + 5}
          map_1_A(x1) = max{4, x1 + 2}
          app#!_A(x1,x2) = max{8, x1 - 6, x2 - 1}
          inc_0!_A = 1
          inc_1#!_A(x1) = max{7, x1 - 2}
          map_2#!_A(x1,x2) = max{6, x1 - 5, x2 - 6}
          plus_1!_A(x1) = 2
          s_1!_A(x1) = max{2, x1 - 15}
          |0|_0!_A = 0
          cons_2!_A(x1,x2) = max{3, x1 - 13, x2 - 17}
          map_1!_A(x1) = 5
  
    partial status:
  
      pi(app#) = []
      pi(inc_0) = []
      pi(inc_1#) = []
      pi(map_2#) = []
      pi(plus_1) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(cons_2) = []
      pi(map_1) = [1]

The following rules are removed:

  p1, p2, p4, p5



-- 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(inc_0(),y) -> inc_1(y)
r4: app(map_0(),y) -> map_1(y)
r5: app(map_1(x1),y) -> map_2(x1,y)
r6: app(plus_0(),y) -> plus_1(y)
r7: app(plus_1(x1),y) -> plus_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: plus_2(|0|_0(),y) -> y
r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
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))

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(inc_0(),y) -> inc_1(y)
r4: app(map_0(),y) -> map_1(y)
r5: app(map_1(x1),y) -> map_2(x1,y)
r6: app(plus_0(),y) -> plus_1(y)
r7: app(plus_1(x1),y) -> plus_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: plus_2(|0|_0(),y) -> y
r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
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))

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{9, x2}
        cons_2_A(x1,x2) = max{x1 + 3, 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{4, x1 - 4, x2 - 4}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{1, x1 - 14, x2 - 23}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 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: plus_2#(s_1(x),y) -> plus_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(inc_0(),y) -> inc_1(y)
r4: app(map_0(),y) -> map_1(y)
r5: app(map_1(x1),y) -> map_2(x1,y)
r6: app(plus_0(),y) -> plus_1(y)
r7: app(plus_1(x1),y) -> plus_2(x1,y)
r8: app(s_0(),y) -> s_1(y)
r9: plus_2(|0|_0(),y) -> y
r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
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))

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:
      
        plus_2#_A(x1,x2) = max{x1 + 7, x2 + 8}
        s_1_A(x1) = x1 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus_2#_A(x1,x2) = 0
          s_1_A(x1) = max{8, x1 - 8}
          plus_2#!_A(x1,x2) = 0
          s_1!_A(x1) = max{0, x1 - 30}
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.