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))

-- SCC decomposition.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(s(),app(app(plus(),x),y))
p2: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)
p3: app#(app(plus(),app(s(),x)),y) -> app#(plus(),x)
p4: app#(app(times(),app(s(),x)),y) -> app#(app(plus(),app(app(times(),x),y)),y)
p5: app#(app(times(),app(s(),x)),y) -> app#(plus(),app(app(times(),x),y))
p6: app#(app(times(),app(s(),x)),y) -> app#(app(times(),x),y)
p7: app#(app(times(),app(s(),x)),y) -> app#(times(),x)
p8: app#(inc(),xs) -> app#(app(map(),app(plus(),app(s(),|0|()))),xs)
p9: app#(inc(),xs) -> app#(map(),app(plus(),app(s(),|0|())))
p10: app#(inc(),xs) -> app#(plus(),app(s(),|0|()))
p11: app#(inc(),xs) -> app#(s(),|0|())
p12: app#(double(),xs) -> app#(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
p13: app#(double(),xs) -> app#(map(),app(times(),app(s(),app(s(),|0|()))))
p14: app#(double(),xs) -> app#(times(),app(s(),app(s(),|0|())))
p15: app#(double(),xs) -> app#(s(),app(s(),|0|()))
p16: app#(double(),xs) -> app#(s(),|0|())
p17: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p18: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p19: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p20: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(times(),|0|()),y) -> |0|()
r4: app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
r5: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
r6: app(double(),xs) -> app(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
r7: app(app(map(),f),nil()) -> nil()
r8: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p8, p12, p19, p20}
  {p6}
  {p2}


-- Reduction pair.

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

p1: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p3: app#(double(),xs) -> app#(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
p4: app#(inc(),xs) -> app#(app(map(),app(plus(),app(s(),|0|()))),xs)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(times(),|0|()),y) -> |0|()
r4: app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
r5: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
r6: app(double(),xs) -> app(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
r7: app(app(map(),f),nil()) -> nil()
r8: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x2
      app_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,0,1)) x2 + (1,1,1,0)
      map_A() = (1,1,1,1)
      cons_A() = (1,1,1,0)
      double_A() = (10,1,16,1)
      times_A() = (1,1,1,1)
      s_A() = (1,1,1,1)
      |0|_A() = (1,1,1,0)
      inc_A() = (5,5,9,1)
      plus_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(app(times(),app(s(),x)),y) -> app#(app(times(),x),y)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(times(),|0|()),y) -> |0|()
r4: app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
r5: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
r6: app(double(),xs) -> app(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
r7: app(app(map(),f),nil()) -> nil()
r8: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2
      app_A(x1,x2) = x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,1)) x2 + (0,0,1,0)
      times_A() = (1,1,1,1)
      s_A() = (0,0,1,1)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(times(),|0|()),y) -> |0|()
r4: app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
r5: app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs)
r6: app(double(),xs) -> app(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
r7: app(app(map(),f),nil()) -> nil()
r8: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2
      app_A(x1,x2) = x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,1)) x2 + (0,0,1,0)
      plus_A() = (1,1,1,1)
      s_A() = (0,0,1,1)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.