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(app(times(),|0|()),y) -> |0|()
  app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
  app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
  app(double(),xs) -> app(app(map(),app(times(),app(s(),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(double_0(),y) -> double_1(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)
  app(times_0(),y) -> times_1(y)
  app(times_1(x1),y) -> times_2(x1,y)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  times_2(|0|_0(),y) -> |0|_0()
  times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
  inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
  double_1(xs) -> map_2(times_1(s_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#(double_0(),y) -> double_1#(y)
p2: app#(inc_0(),y) -> inc_1#(y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(plus_1(x1),y) -> plus_2#(x1,y)
p5: app#(times_1(x1),y) -> times_2#(x1,y)
p6: plus_2#(s_1(x),y) -> plus_2#(x,y)
p7: times_2#(s_1(x),y) -> plus_2#(times_2(x,y),y)
p8: times_2#(s_1(x),y) -> times_2#(x,y)
p9: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs)
p10: double_1#(xs) -> map_2#(times_1(s_1(s_1(|0|_0()))),xs)
p11: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: 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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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, p3, p9, p10, p11, p12}
  {p8}
  {p6}


-- Reduction pair.

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

p1: app#(double_0(),y) -> double_1#(y)
p2: double_1#(xs) -> map_2#(times_1(s_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)
p6: app#(inc_0(),y) -> inc_1#(y)
p7: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1 + 4
          double_0_A() = 13
          double_1#_A(x1) = 12
          map_2#_A(x1,x2) = x1 + 5
          times_1_A(x1) = x1 + 4
          s_1_A(x1) = 2
          |0|_0_A() = 13
          cons_2_A(x1,x2) = 3
          map_1_A(x1) = x1 + 6
          inc_0_A() = 23
          inc_1#_A(x1) = 22
          plus_1_A(x1) = 16
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = 0
            double_0_A() = 1
            double_1#_A(x1) = 0
            map_2#_A(x1,x2) = 0
            times_1_A(x1) = 2
            s_1_A(x1) = 2
            |0|_0_A() = 5
            cons_2_A(x1,x2) = 4
            map_1_A(x1) = 2
            inc_0_A() = 1
            inc_1#_A(x1) = 0
            plus_1_A(x1) = 2
            app#!_A(x1,x2) = 12
            double_0!_A() = 0
            double_1#!_A(x1) = 4
            map_2#!_A(x1,x2) = 10
            times_1!_A(x1) = x1 + 3
            s_1!_A(x1) = 6
            |0|_0!_A() = 6
            cons_2!_A(x1,x2) = 9
            map_1!_A(x1) = 11
            inc_0!_A() = 10
            inc_1#!_A(x1) = x1 + 11
            plus_1!_A(x1) = 2
  
    partial status:
  
      pi(app#) = []
      pi(double_0) = []
      pi(double_1#) = []
      pi(map_2#) = []
      pi(times_1) = [1]
      pi(s_1) = []
      pi(|0|_0) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(inc_0) = []
      pi(inc_1#) = []
      pi(plus_1) = []

The following rules are removed:

  p1, p2, p4, p5, p6, p7



-- 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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          map_2#_A(x1,x2) = 1
          cons_2_A(x1,x2) = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            map_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = x2 + 2
            map_2#!_A(x1,x2) = x2
            cons_2!_A(x1,x2) = 7
  
    partial status:
  
      pi(map_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: times_2#(s_1(x),y) -> times_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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          times_2#_A(x1,x2) = x1
          s_1_A(x1) = x1 + 15
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            times_2#_A(x1,x2) = x1
            s_1_A(x1) = x1 + 2
            times_2#!_A(x1,x2) = x1 + 5
            s_1!_A(x1) = 4
  
    partial status:
  
      pi(times_2#) = []
      pi(s_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: 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(double_0(),y) -> double_1(y)
r4: app(inc_0(),y) -> inc_1(y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(times_0(),y) -> times_1(y)
r11: app(times_1(x1),y) -> times_2(x1,y)
r12: plus_2(|0|_0(),y) -> y
r13: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r14: times_2(|0|_0(),y) -> |0|_0()
r15: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r16: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs)
r17: double_1(xs) -> map_2(times_1(s_1(s_1(|0|_0()))),xs)
r18: map_2(f,nil_0()) -> nil_0()
r19: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus_2#_A(x1,x2) = x1
          s_1_A(x1) = x1 + 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            plus_2#_A(x1,x2) = x1
            s_1_A(x1) = x1 + 2
            plus_2#!_A(x1,x2) = x1 + 3
            s_1!_A(x1) = 2
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.