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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,1),(0,1)) x1 + (8,6)
          app_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,0),(0,1)) x2 + (0,2)
          map_A() = (19,0)
          cons_A() = (20,1)
          double_A() = (2,10)
          times_A() = (1,17)
          s_A() = (21,21)
          |0|_A() = (19,1)
          inc_A() = (1,12)
          plus_A() = (21,21)
  
    precedence:
    
      app# = app = map = cons = double = times = s = |0| = inc = plus
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(map) = []
      pi(cons) = []
      pi(double) = []
      pi(times) = []
      pi(s) = []
      pi(|0|) = []
      pi(inc) = []
      pi(plus) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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)

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:

  {p1, p2, p3}


-- 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#(double(),xs) -> app#(app(map(),app(times(),app(s(),app(s(),|0|())))),xs)
p3: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + (19,4)
          app_A(x1,x2) = ((1,1),(0,0)) x2 + (1,2)
          map_A() = (18,3)
          cons_A() = (2,1)
          double_A() = (17,1)
          times_A() = (18,2)
          s_A() = (1,5)
          |0|_A() = (1,5)
  
    precedence:
    
      app# = app = map = cons = double = times = s = |0|
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(map) = []
      pi(cons) = []
      pi(double) = []
      pi(times) = []
      pi(s) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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)

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:

  {p1, 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)

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,1)) x1 + (3,3)
          app_A(x1,x2) = x2 + (2,2)
          map_A() = (0,0)
          cons_A() = (1,1)
  
    precedence:
    
      app# = app = cons > map
    
    partial status:
  
      pi(app#) = []
      pi(app) = [2]
      pi(map) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  {p1}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,0)) x2 + (4,3)
          app_A(x1,x2) = ((1,0),(0,0)) x2 + (2,2)
          map_A() = (1,1)
          cons_A() = (3,2)
  
    precedence:
    
      app# = app = map = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(map) = []
      pi(cons) = []

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(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,6)
          app_A(x1,x2) = x2 + (1,2)
          times_A() = (5,5)
          s_A() = (6,1)
  
    precedence:
    
      app# > times > app > s
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(times) = []
      pi(s) = []

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,6)
          app_A(x1,x2) = x2 + (1,2)
          plus_A() = (3,5)
          s_A() = (4,1)
  
    precedence:
    
      app# = app > plus > s
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(plus) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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