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)))

Instead we prove termination of the uncurried system:

  app(append_0(),y) -> append_1(y)
  app(append_1(x1),y) -> append_2(x1,y)
  app(combine_0(),y) -> combine_1(y)
  app(combine_1(x1),y) -> combine_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(levels_0(),y) -> levels_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(node_0(),y) -> node_1(y)
  app(node_1(x1),y) -> node_2(x1,y)
  app(zip_0(),y) -> zip_1(y)
  app(zip_1(x1),y) -> zip_2(x1,y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  append_2(xs,nil_0()) -> xs
  append_2(nil_0(),ys) -> ys
  append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
  zip_2(nil_0(),yss) -> yss
  zip_2(xss,nil_0()) -> xss
  zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
  combine_2(xs,nil_0()) -> xs
  combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
  levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))


-- SCC decomposition.

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

p1: app#(append_1(x1),y) -> append_2#(x1,y)
p2: app#(combine_1(x1),y) -> combine_2#(x1,y)
p3: app#(levels_0(),y) -> levels_1#(y)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: app#(zip_1(x1),y) -> zip_2#(x1,y)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p8: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)
p9: zip_2#(cons_2(xs,xss),cons_2(ys,yss)) -> append_2#(xs,ys)
p10: zip_2#(cons_2(xs,xss),cons_2(ys,yss)) -> zip_2#(xss,yss)
p11: combine_2#(xs,cons_2(ys,yss)) -> combine_2#(zip_2(xs,ys),yss)
p12: combine_2#(xs,cons_2(ys,yss)) -> zip_2#(xs,ys)
p13: levels_1#(node_2(x,xs)) -> combine_2#(nil_0(),map_2(levels_0(),xs))
p14: levels_1#(node_2(x,xs)) -> map_2#(levels_0(),xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(combine_0(),y) -> combine_1(y)
r4: app(combine_1(x1),y) -> combine_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(levels_0(),y) -> levels_1(y)
r8: app(map_0(),y) -> map_1(y)
r9: app(map_1(x1),y) -> map_2(x1,y)
r10: app(node_0(),y) -> node_1(y)
r11: app(node_1(x1),y) -> node_2(x1,y)
r12: app(zip_0(),y) -> zip_1(y)
r13: app(zip_1(x1),y) -> zip_2(x1,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: append_2(xs,nil_0()) -> xs
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: zip_2(nil_0(),yss) -> yss
r20: zip_2(xss,nil_0()) -> xss
r21: zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
r22: combine_2(xs,nil_0()) -> xs
r23: combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
r24: levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p6, p7, p14}
  {p11}
  {p10}
  {p8}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: app#(levels_0(),y) -> levels_1#(y)
p5: levels_1#(node_2(x,xs)) -> map_2#(levels_0(),xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(combine_0(),y) -> combine_1(y)
r4: app(combine_1(x1),y) -> combine_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(levels_0(),y) -> levels_1(y)
r8: app(map_0(),y) -> map_1(y)
r9: app(map_1(x1),y) -> map_2(x1,y)
r10: app(node_0(),y) -> node_1(y)
r11: app(node_1(x1),y) -> node_2(x1,y)
r12: app(zip_0(),y) -> zip_1(y)
r13: app(zip_1(x1),y) -> zip_2(x1,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: append_2(xs,nil_0()) -> xs
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: zip_2(nil_0(),yss) -> yss
r20: zip_2(xss,nil_0()) -> xss
r21: zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
r22: combine_2(xs,nil_0()) -> xs
r23: combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
r24: levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 8
              app#_A(x1,x2) = x2
              map_1_A(x1) = 3
              levels_0_A() = 7
              levels_1#_A(x1) = x1
              node_2_A(x1,x2) = x1 + x2 + 11
              map_2#!_A(x1,x2) = x2 + 9
              cons_2!_A(x1,x2) = 5
              app#!_A(x1,x2) = x2 + 10
              map_1!_A(x1) = 0
              levels_0!_A() = 7
              levels_1#!_A(x1) = x1
              node_2!_A(x1,x2) = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x2
              map_1_A(x1) = 3
              levels_0_A() = 1
              levels_1#_A(x1) = x1
              node_2_A(x1,x2) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x2 + 11
              cons_2!_A(x1,x2) = 4
              app#!_A(x1,x2) = x2 + 12
              map_1!_A(x1) = 10
              levels_0!_A() = 8
              levels_1#!_A(x1) = 9
              node_2!_A(x1,x2) = 7
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(levels_0) = []
      pi(levels_1#) = []
      pi(node_2) = []

The following rules are removed:

  p1, p2, p3, p4, p5

Then no dependency pair remains.

-- Reduction pair.

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

p1: combine_2#(xs,cons_2(ys,yss)) -> combine_2#(zip_2(xs,ys),yss)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(combine_0(),y) -> combine_1(y)
r4: app(combine_1(x1),y) -> combine_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(levels_0(),y) -> levels_1(y)
r8: app(map_0(),y) -> map_1(y)
r9: app(map_1(x1),y) -> map_2(x1,y)
r10: app(node_0(),y) -> node_1(y)
r11: app(node_1(x1),y) -> node_2(x1,y)
r12: app(zip_0(),y) -> zip_1(y)
r13: app(zip_1(x1),y) -> zip_2(x1,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: append_2(xs,nil_0()) -> xs
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: zip_2(nil_0(),yss) -> yss
r20: zip_2(xss,nil_0()) -> xss
r21: zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
r22: combine_2(xs,nil_0()) -> xs
r23: combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
r24: levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))

The set of usable rules consists of

  r16, r17, r18, r19, r20, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              combine_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              zip_2_A(x1,x2) = x1 + x2 + 5
              append_2_A(x1,x2) = x1 + x2 + 5
              nil_0_A() = 7
              combine_2#!_A(x1,x2) = 6
              cons_2!_A(x1,x2) = 7
              zip_2!_A(x1,x2) = x1 + x2 + 12
              append_2!_A(x1,x2) = 6
              nil_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              combine_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 9
              zip_2_A(x1,x2) = x1 + x2 + 9
              append_2_A(x1,x2) = x1 + x2 + 9
              nil_0_A() = 6
              combine_2#!_A(x1,x2) = x2
              cons_2!_A(x1,x2) = x1 + 39
              zip_2!_A(x1,x2) = x1 + x2 + 10
              append_2!_A(x1,x2) = x1 + 29
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(combine_2#) = []
      pi(cons_2) = []
      pi(zip_2) = [1, 2]
      pi(append_2) = [1, 2]
      pi(nil_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: zip_2#(cons_2(xs,xss),cons_2(ys,yss)) -> zip_2#(xss,yss)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(combine_0(),y) -> combine_1(y)
r4: app(combine_1(x1),y) -> combine_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(levels_0(),y) -> levels_1(y)
r8: app(map_0(),y) -> map_1(y)
r9: app(map_1(x1),y) -> map_2(x1,y)
r10: app(node_0(),y) -> node_1(y)
r11: app(node_1(x1),y) -> node_2(x1,y)
r12: app(zip_0(),y) -> zip_1(y)
r13: app(zip_1(x1),y) -> zip_2(x1,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: append_2(xs,nil_0()) -> xs
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: zip_2(nil_0(),yss) -> yss
r20: zip_2(xss,nil_0()) -> xss
r21: zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
r22: combine_2(xs,nil_0()) -> xs
r23: combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
r24: levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              zip_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 5
              zip_2#!_A(x1,x2) = 0
              cons_2!_A(x1,x2) = x1 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              zip_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x2 + 9
              zip_2#!_A(x1,x2) = x2 + 11
              cons_2!_A(x1,x2) = 12
        
  
    partial status:
  
      pi(zip_2#) = [1]
      pi(cons_2) = [2]

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: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(combine_0(),y) -> combine_1(y)
r4: app(combine_1(x1),y) -> combine_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(levels_0(),y) -> levels_1(y)
r8: app(map_0(),y) -> map_1(y)
r9: app(map_1(x1),y) -> map_2(x1,y)
r10: app(node_0(),y) -> node_1(y)
r11: app(node_1(x1),y) -> node_2(x1,y)
r12: app(zip_0(),y) -> zip_1(y)
r13: app(zip_1(x1),y) -> zip_2(x1,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: append_2(xs,nil_0()) -> xs
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: zip_2(nil_0(),yss) -> yss
r20: zip_2(xss,nil_0()) -> xss
r21: zip_2(cons_2(xs,xss),cons_2(ys,yss)) -> cons_2(append_2(xs,ys),zip_2(xss,yss))
r22: combine_2(xs,nil_0()) -> xs
r23: combine_2(xs,cons_2(ys,yss)) -> combine_2(zip_2(xs,ys),yss)
r24: levels_1(node_2(x,xs)) -> cons_2(cons_2(x,nil_0()),combine_2(nil_0(),map_2(levels_0(),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              append_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 5
              append_2#!_A(x1,x2) = 0
              cons_2!_A(x1,x2) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              append_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x2 + 5
              append_2#!_A(x1,x2) = x1 + 1
              cons_2!_A(x1,x2) = x2 + 2
        
  
    partial status:
  
      pi(append_2#) = []
      pi(cons_2) = []

The following rules are removed:

  p1

Then no dependency pair remains.