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:
  
      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:
          
            app#_A(x1,x2) = max{0, x1 - 17, x2 - 15}
            inc_0_A = 14
            inc_1#_A(x1) = max{0, x1 - 15}
            map_2#_A(x1,x2) = max{0, x1 - 14, x2 - 23}
            plus_1_A(x1) = max{11, x1 + 2}
            s_1_A(x1) = max{11, x1 - 10}
            |0|_0_A = 8
            cons_2_A(x1,x2) = max{x1 + 22, x2 + 22}
            map_1_A(x1) = x1 + 16
            app#!_A(x1,x2) = max{5, x1 - 15, x2 - 1}
            inc_0!_A = 4
            inc_1#!_A(x1) = max{5, x1 - 15}
            map_2#!_A(x1,x2) = max{4, x1 - 14, x2 - 16}
            plus_1!_A(x1) = max{3, x1 - 12}
            s_1!_A(x1) = max{6, x1 - 33}
            |0|_0!_A = 6
            cons_2!_A(x1,x2) = max{1, x1 - 3, x2 - 24}
            map_1!_A(x1) = max{1, x1 - 37}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 28, x2 - 20}
            inc_0_A = 27
            inc_1#_A(x1) = max{0, x1 - 20}
            map_2#_A(x1,x2) = max{0, x1 - 26, x2 - 20}
            plus_1_A(x1) = x1 + 16
            s_1_A(x1) = max{9, x1 - 9}
            |0|_0_A = 0
            cons_2_A(x1,x2) = max{x1 + 13, x2 + 15}
            map_1_A(x1) = max{27, x1 + 2}
            app#!_A(x1,x2) = max{5, x1 - 30, x2 + 3}
            inc_0!_A = 5
            inc_1#!_A(x1) = max{4, x1 - 25}
            map_2#!_A(x1,x2) = max{2, x1 - 29, x2 - 26}
            plus_1!_A(x1) = max{6, x1 - 35}
            s_1!_A(x1) = max{7, x1 - 24}
            |0|_0!_A = 7
            cons_2!_A(x1,x2) = x1 + 1
            map_1!_A(x1) = max{1, x1 - 29}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(inc_0) = []
      pi(inc_1#) = []
      pi(map_2#) = [1]
      pi(plus_1) = []
      pi(s_1) = [1]
      pi(|0|_0) = []
      pi(cons_2) = [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:
  
      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) = 0
            cons_2_A(x1,x2) = max{13, x1 - 2, x2 - 4}
            map_2#!_A(x1,x2) = max{0, x2 - 14}
            cons_2!_A(x1,x2) = max{2, x1 - 3, x2 - 13}
        
        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 - 2, x2 - 4}
            map_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 5}
            cons_2!_A(x1,x2) = max{0, x1 - 11}
        
  
    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:
  
      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:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{10, x1 - 3}
            plus_2#!_A(x1,x2) = max{0, x1 - 11}
            s_1!_A(x1) = max{6, x1 - 6}
        
        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{9, x1 - 3}
            plus_2#!_A(x1,x2) = max{7, x1 - 10, x2 - 1}
            s_1!_A(x1) = max{6, x1 - 16}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.