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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(1,1,0),(1,0,0)) x1 + ((1,0,0),(1,1,0),(0,0,0)) x2
      app_A(x1,x2) = x1 + ((1,0,0),(1,1,0),(0,0,0)) x2 + (0,0,1)
      map_A() = (1,2,1)
      cons_A() = (1,1,0)
      levels_A() = (1,1,1)
      node_A() = (2,1,0)

The next rules are strictly ordered:

  p1, p2, p3

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x1 + x2
      app_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(0,0,0),(1,0,0)) x2 + (1,1,0)
      combine_A() = (0,1,1)
      cons_A() = (2,1,0)
      zip_A() = (1,1,0)
      append_A() = (1,1,0)
      nil_A() = (1,1,0)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      app_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
      zip_A() = (1,1,1)
      cons_A() = (1,1,0)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      app_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,1)) x2 + (1,1,1)
      append_A() = (1,1,1)
      cons_A() = (1,1,0)

The next rules are strictly ordered:

  p1

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