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(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))
  inc() -> app(map(),app(plus(),app(s(),|0|())))
  double() -> app(map(),app(times(),app(s(),app(s(),|0|()))))

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(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)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  inc_0() -> map_1(plus_1(s_1(|0|_0())))
  double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))


-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: app#(plus_1(x1),y) -> plus_2#(x1,y)
p3: app#(times_1(x1),y) -> times_2#(x1,y)
p4: plus_2#(s_1(x),y) -> plus_2#(x,y)
p5: times_2#(s_1(x),y) -> plus_2#(times_2(x,y),y)
p6: times_2#(s_1(x),y) -> times_2#(x,y)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: 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(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The estimated dependency graph contains the following SCCs:

  {p1, p7, p8}
  {p6}
  {p4}


-- Reduction pair.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: map_2#(f,cons_2(x,xs)) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r16, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x2 + 5
        map_1_A(x1) = max{2, x1 - 11}
        map_2#_A(x1,x2) = max{3, x2 - 2}
        cons_2_A(x1,x2) = max{x1 + 8, x2 + 1}
        inc_0_A = 11
        plus_1_A(x1) = max{10, x1 - 14}
        s_1_A(x1) = max{2, x1 - 12}
        |0|_0_A = 0
        double_0_A = 11
        times_1_A(x1) = max{10, x1 - 5}
  
    base order 2:
  
      precedence:
      
        inc_0 = double_0 > |0|_0 = times_1 > map_1 = plus_1 = s_1 > cons_2 > map_2# > app#
      
  
    partial status:
  
      pi(app#) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(cons_2) = [1]
      pi(inc_0) = []
      pi(plus_1) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(double_0) = []
      pi(times_1) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r3, r5, r7, r8, r16, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{11, x2 + 10}
        map_1_A(x1) = max{2, x1 - 11}
        map_2#_A(x1,x2) = max{11, x2 + 1}
        cons_2_A(x1,x2) = max{x1 + 9, x2 - 12}
        app_A(x1,x2) = max{7, x1 - 5, x2 - 10}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 11}
        map_0_A = 1
        plus_0_A = 3
        plus_1_A(x1) = max{4, x1 - 11}
        s_0_A = 1
        s_1_A(x1) = max{2, x1 - 18}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 11}
        inc_0_A = 3
        |0|_0_A = 0
        double_0_A = 5
  
    base order 2:
  
      precedence:
      
        app# = map_2# = app > cons_2 = double_0 > cons_1 = times_1 > cons_0 = plus_1 = times_0 > plus_0 = inc_0 > map_1 = s_1 = |0|_0 > map_0 = s_0
      
  
    partial status:
  
      pi(app#) = [2]
      pi(map_1) = []
      pi(map_2#) = [2]
      pi(cons_2) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(map_0) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(inc_0) = []
      pi(|0|_0) = []
      pi(double_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: 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(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r2, r3, r5, r8, r16, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = x1 + 1
        s_1_A(x1) = x1 + 3
        app_A(x1,x2) = max{6, x1 - 7, x2 - 3}
        cons_0_A = 5
        cons_1_A(x1) = max{6, x1 - 3}
        cons_2_A(x1,x2) = max{1, x1 - 13, x2 - 6}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 4}
        plus_0_A = 8
        plus_1_A(x1) = max{3, x1 - 4}
        times_0_A = 8
        times_1_A(x1) = max{3, x1 - 7}
        inc_0_A = 4
        |0|_0_A = 0
        double_0_A = 7
  
    base order 2:
  
      precedence:
      
        app > cons_0 = inc_0 = double_0 > s_1 = cons_1 = cons_2 = map_1 = plus_1 = times_1 = |0|_0 > times_2# = map_0 = plus_0 = times_0
      
  
    partial status:
  
      pi(times_2#) = [1]
      pi(s_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(inc_0) = []
      pi(|0|_0) = []
      pi(double_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: 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(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(plus_0(),y) -> plus_1(y)
r6: app(plus_1(x1),y) -> plus_2(x1,y)
r7: app(s_0(),y) -> s_1(y)
r8: app(times_0(),y) -> times_1(y)
r9: app(times_1(x1),y) -> times_2(x1,y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: inc_0() -> map_1(plus_1(s_1(|0|_0())))
r17: double_0() -> map_1(times_1(s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r2, r3, r5, r8, r16, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = max{4, x1 + 1}
        s_1_A(x1) = max{10, x1 + 5}
        app_A(x1,x2) = max{4, x1 - 11, x2 - 11}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 11}
        cons_2_A(x1,x2) = max{1, x1 - 23, x2 - 12}
        map_0_A = 14
        map_1_A(x1) = max{2, x1 - 13}
        plus_0_A = 23
        plus_1_A(x1) = max{12, x1 - 11}
        times_0_A = 10
        times_1_A(x1) = max{3, x1 - 16}
        inc_0_A = 13
        |0|_0_A = 0
        double_0_A = 16
  
    base order 2:
  
      precedence:
      
        app = inc_0 = double_0 > map_1 > times_0 > s_1 = cons_2 = map_0 = plus_1 = times_1 = |0|_0 > plus_2# = cons_1 = plus_0 > cons_0
      
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(inc_0) = []
      pi(|0|_0) = []
      pi(double_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.