YES

We show the termination of the TRS R:

  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))
  app(app(append(),xs),nil()) -> xs
  app(app(append(),nil()),ys) -> ys
  app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
  app(app(zip(),nil()),yss) -> yss
  app(app(zip(),xss),nil()) -> xss
  app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
  app(app(combine(),xs),nil()) -> xs
  app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
  app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

-- 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(cons(),app(f,x)),app(app(map(),f),xs))
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p5: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(cons(),x),app(app(append(),xs),ys))
p6: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(append(),xs),ys)
p7: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(append(),xs)
p8: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
p9: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(cons(),app(app(append(),xs),ys))
p10: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(app(append(),xs),ys)
p11: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(append(),xs)
p12: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(app(zip(),xss),yss)
p13: app#(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(zip(),xss)
p14: app#(app(combine(),xs),app(app(cons(),ys),yss)) -> app#(app(combine(),app(app(zip(),xs),ys)),yss)
p15: app#(app(combine(),xs),app(app(cons(),ys),yss)) -> app#(combine(),app(app(zip(),xs),ys))
p16: app#(app(combine(),xs),app(app(cons(),ys),yss)) -> app#(app(zip(),xs),ys)
p17: app#(app(combine(),xs),app(app(cons(),ys),yss)) -> app#(zip(),xs)
p18: app#(levels(),app(app(node(),x),xs)) -> app#(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))
p19: app#(levels(),app(app(node(),x),xs)) -> app#(cons(),app(app(cons(),x),nil()))
p20: app#(levels(),app(app(node(),x),xs)) -> app#(app(cons(),x),nil())
p21: app#(levels(),app(app(node(),x),xs)) -> app#(cons(),x)
p22: app#(levels(),app(app(node(),x),xs)) -> app#(app(combine(),nil()),app(app(map(),levels()),xs))
p23: app#(levels(),app(app(node(),x),xs)) -> app#(combine(),nil())
p24: app#(levels(),app(app(node(),x),xs)) -> app#(app(map(),levels()),xs)
p25: app#(levels(),app(app(node(),x),xs)) -> app#(map(),levels())

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p24}
  {p14}
  {p12}
  {p6}


-- 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#(levels(),app(app(node(),x),xs)) -> app#(app(map(),levels()),xs)
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{20, x1, x2 + 10}
          app_A(x1,x2) = max{22, x1 + 3, x2 + 16}
          map_A = 21
          cons_A = 22
          levels_A = 9
          node_A = 7
    
      precedence:
      
        app = map = cons > levels = node > app#
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
        pi(levels) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 10, x2 + 8}
          app_A(x1,x2) = max{5, x1, x2}
          map_A = 1
          cons_A = 4
          levels_A = 2
          node_A = 6
    
      precedence:
      
        app# = app > map = cons > levels = node
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
        pi(levels) = []
        pi(node) = []
    

The next rules are strictly ordered:

  p1, 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#(app(map(),f),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),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(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x2 - 4}
          app_A(x1,x2) = max{x1 + 6, x2 + 3}
          map_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = max{x1 + 2, x2 + 6}
          map_A = 3
          cons_A = 4
    
      precedence:
      
        app# = app = map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        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(combine(),xs),app(app(cons(),ys),yss)) -> app#(app(combine(),app(app(zip(),xs),ys)),yss)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The set of usable rules consists of

  r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (10,1)
            app_A(x1,x2) = x1 + x2 + (3,3)
            combine_A() = (2,4)
            cons_A() = (3,6)
            zip_A() = (1,5)
            append_A() = (2,4)
            nil_A() = (1,0)
    
      precedence:
      
        app# = cons > app = combine = zip = append > nil
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(combine) = []
        pi(cons) = []
        pi(zip) = []
        pi(append) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (9,9)
            app_A(x1,x2) = x1 + (4,4)
            combine_A() = (0,0)
            cons_A() = (6,6)
            zip_A() = (0,0)
            append_A() = (3,3)
            nil_A() = (1,0)
    
      precedence:
      
        combine = cons = zip > app# > app = append = nil
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1]
        pi(combine) = []
        pi(cons) = []
        pi(zip) = []
        pi(append) = []
        pi(nil) = []
    

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(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app#(app(zip(),xss),yss)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 4, x2 + 8}
          app_A(x1,x2) = max{x1 + 2, x2 + 4}
          zip_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = zip = cons
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(zip) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 15, x2 - 8}
          app_A(x1,x2) = max{x1 + 5, x2 + 7}
          zip_A = 2
          cons_A = 1
    
      precedence:
      
        app# = app = zip = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(zip) = []
        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(append(),app(app(cons(),x),xs)),ys) -> app#(app(append(),xs),ys)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(append(),xs),nil()) -> xs
r4: app(app(append(),nil()),ys) -> ys
r5: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r6: app(app(zip(),nil()),yss) -> yss
r7: app(app(zip(),xss),nil()) -> xss
r8: app(app(zip(),app(app(cons(),xs),xss)),app(app(cons(),ys),yss)) -> app(app(cons(),app(app(append(),xs),ys)),app(app(zip(),xss),yss))
r9: app(app(combine(),xs),nil()) -> xs
r10: app(app(combine(),xs),app(app(cons(),ys),yss)) -> app(app(combine(),app(app(zip(),xs),ys)),yss)
r11: app(levels(),app(app(node(),x),xs)) -> app(app(cons(),app(app(cons(),x),nil())),app(app(combine(),nil()),app(app(map(),levels()),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 4, x2 + 5}
          app_A(x1,x2) = max{x1 + 2, x2 + 4}
          append_A = 2
          cons_A = 3
    
      precedence:
      
        app = append > app# = cons
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(append) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 - 15, x2 + 3}
          app_A(x1,x2) = max{x1 + 4, x2 + 7}
          append_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = append = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(append) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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