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(treemap(),f),app(app(node(),x),xs)) -> app(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,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(treemap_0(),y) -> treemap_1(y)
  app(treemap_1(x1),y) -> treemap_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))
  treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs))


-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: app#(treemap_1(x1),y) -> treemap_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: treemap_2#(f,node_2(x,xs)) -> app#(f,x)
p6: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(treemap_0(),y) -> treemap_1(y)
r8: app(treemap_1(x1),y) -> treemap_2(x1,y)
r9: map_2(f,nil_0()) -> nil_0()
r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p4: app#(treemap_1(x1),y) -> treemap_2#(x1,y)
p5: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs)
p6: treemap_2#(f,node_2(x,xs)) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(treemap_0(),y) -> treemap_1(y)
r8: app(treemap_1(x1),y) -> treemap_2(x1,y)
r9: map_2(f,nil_0()) -> nil_0()
r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1 + x2 + 8
          map_1_A(x1) = x1 + 2
          map_2#_A(x1,x2) = x1 + x2 + 4
          cons_2_A(x1,x2) = x1 + x2 + 5
          treemap_1_A(x1) = x1 + 8
          treemap_2#_A(x1,x2) = x1 + x2 + 9
          node_2_A(x1,x2) = x1 + x2 + 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 5, x2 - 9}
          map_1_A(x1) = x1 + 4
          map_2#_A(x1,x2) = max{0, x1 - 5, x2 - 22}
          cons_2_A(x1,x2) = max{14, x1 + 13, x2 + 13}
          treemap_1_A(x1) = max{2, x1 + 1}
          treemap_2#_A(x1,x2) = max{0, x1 - 4, x2 - 9}
          node_2_A(x1,x2) = max{x1 + 5, x2 - 3}
          app#!_A(x1,x2) = max{4, x1 - 11, x2 - 13}
          map_1!_A(x1) = max{3, x1 + 1}
          map_2#!_A(x1,x2) = max{5, x1 - 10, x2 - 15}
          cons_2!_A(x1,x2) = max{3, x1 - 2, x2 - 13}
          treemap_1!_A(x1) = max{2, x1 - 10}
          treemap_2#!_A(x1,x2) = max{5, x1 - 9, x2 - 12}
          node_2!_A(x1,x2) = max{0, x1 - 15, x2 - 13}
  
    partial status:
  
      pi(app#) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(cons_2) = [2]
      pi(treemap_1) = []
      pi(treemap_2#) = [2]
      pi(node_2) = [2]

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.