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:
  
      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 - 62, x2 - 10}
            double_0_A = 0
            double_1#_A(x1) = max{0, x1 - 47}
            map_2#_A(x1,x2) = max{0, x1 - 61, x2 - 57}
            times_1_A(x1) = x1 + 41
            s_1_A(x1) = max{8, x1 - 16}
            |0|_0_A = 6
            cons_2_A(x1,x2) = max{x1 + 47, x2 + 43}
            map_1_A(x1) = max{61, x1 + 1}
            inc_0_A = 61
            inc_1#_A(x1) = max{0, x1 - 57}
            plus_1_A(x1) = x1 + 52
            app#!_A(x1,x2) = max{5, x1 - 64, x2 - 3}
            double_0!_A = 6
            double_1#!_A(x1) = max{5, x1 - 50}
            map_2#!_A(x1,x2) = max{5, x1 - 64, x2 - 50}
            times_1!_A(x1) = max{6, x1 - 63}
            s_1!_A(x1) = max{4, x1 - 59}
            |0|_0!_A = 0
            cons_2!_A(x1,x2) = max{1, x1 - 63, x2 - 60}
            map_1!_A(x1) = max{3, x1 - 63}
            inc_0!_A = 6
            inc_1#!_A(x1) = max{5, x1 - 3}
            plus_1!_A(x1) = x1 + 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 6, x2 - 68}
            double_0_A = 5
            double_1#_A(x1) = max{0, x1 - 68}
            map_2#_A(x1,x2) = max{0, x1 - 5, x2 - 82}
            times_1_A(x1) = max{4, x1 - 10}
            s_1_A(x1) = max{14, x1 - 34}
            |0|_0_A = 16
            cons_2_A(x1,x2) = max{x1 + 68, x2 + 81}
            map_1_A(x1) = x1 + 5
            inc_0_A = 5
            inc_1#_A(x1) = max{0, x1 - 68}
            plus_1_A(x1) = max{2, x1 - 12}
            app#!_A(x1,x2) = max{12, x1 - 6, x2 - 2}
            double_0!_A = 1
            double_1#!_A(x1) = max{7, x1 - 68}
            map_2#!_A(x1,x2) = max{6, x1 - 5, x2 - 69}
            times_1!_A(x1) = max{13, x1 - 1}
            s_1!_A(x1) = max{7, x1 - 145}
            |0|_0!_A = 11
            cons_2!_A(x1,x2) = max{3, x1 - 85, x2 - 83}
            map_1!_A(x1) = max{5, x1 - 6}
            inc_0!_A = 9
            inc_1#!_A(x1) = max{11, x1 - 2}
            plus_1!_A(x1) = max{9, x1 - 143}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(double_0) = []
      pi(double_1#) = [1]
      pi(map_2#) = [2]
      pi(times_1) = [1]
      pi(s_1) = []
      pi(|0|_0) = []
      pi(cons_2) = [1, 2]
      pi(map_1) = [1]
      pi(inc_0) = []
      pi(inc_1#) = [1]
      pi(plus_1) = [1]

The following rules are removed:

  p1, p2, p5, 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)
p2: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(inc_0(),y) -> inc_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(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:
  
      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{4, x1 - 11, x2 - 11}
            map_2#!_A(x1,x2) = 1
            cons_2!_A(x1,x2) = max{3, x1 - 18, x2 - 22}
        
        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{34, x1 - 5, x2 - 34}
            map_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 35}
            cons_2!_A(x1,x2) = max{1, x1 - 7, x2 - 39}
        
  
    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: 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:
  
      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:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{5, x1 - 8}
            times_2#!_A(x1,x2) = 1
            s_1!_A(x1) = max{5, x1 - 28}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{8, x1 - 10}
            times_2#!_A(x1,x2) = max{7, x1 - 25, x2 - 1}
            s_1!_A(x1) = max{6, x1 - 25}
        
  
    partial status:
  
      pi(times_2#) = [1]
      pi(s_1) = [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:
  
      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{4, x1 - 6}
            plus_2#!_A(x1,x2) = max{0, x1 - 9}
            s_1!_A(x1) = max{7, x1 - 8}
        
        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{4, x1 - 15}
            plus_2#!_A(x1,x2) = max{4, x1 - 40, x2 - 1}
            s_1!_A(x1) = max{3, x1 - 41}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.