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(app(comp(),f),g),x) -> app(f,app(g,x))
  app(twice(),f) -> app(app(comp(),f),f)

Instead we prove termination of the uncurried system:

  app(comp_0(),y) -> comp_1(y)
  app(comp_1(x1),y) -> comp_2(x1,y)
  app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
  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(twice_0(),y) -> twice_1(y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  comp_3(f,g,x) -> app(f,app(g,x))
  twice_1(f) -> comp_2(f,f)


-- SCC decomposition.

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

p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(twice_0(),y) -> twice_1#(y)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p6: comp_3#(f,g,x) -> app#(f,app(g,x))
p7: comp_3#(f,g,x) -> app#(g,x)

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x))
r12: twice_1(f) -> comp_2(f,f)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5, p6, p7}


-- Reduction pair.

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

p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y)
p2: comp_3#(f,g,x) -> app#(g,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: comp_3#(f,g,x) -> app#(f,app(g,x))

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x))
r12: twice_1(f) -> comp_2(f,f)

The set of usable rules consists of

  r1, r5, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{5, x1}
        comp_2_A(x1,x2) = max{11, x1, x2}
        comp_3#_A(x1,x2,x3) = max{10, x1, x2}
        map_1_A(x1) = max{6, x1}
        map_2#_A(x1,x2) = max{5, x1}
        cons_2_A(x1,x2) = 4
        app_A(x1,x2) = max{10, x1 - 8, x2}
        comp_0_A = 7
        comp_1_A(x1) = max{9, x1 - 1}
        cons_1_A(x1) = max{11, x1}
        map_2_A(x1,x2) = x1 + 11
        nil_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = 1
          comp_2_A(x1,x2) = max{14, x1 - 2, x2}
          comp_3#_A(x1,x2,x3) = 1
          map_1_A(x1) = max{14, x1}
          map_2#_A(x1,x2) = 1
          cons_2_A(x1,x2) = max{23, x1, x2}
          app_A(x1,x2) = max{23, x1 + 6, x2}
          comp_0_A = 0
          comp_1_A(x1) = max{7, x1 - 5}
          cons_1_A(x1) = max{14, x1}
          map_2_A(x1,x2) = max{x1 + 6, x2}
          nil_0_A = 29
          app#!_A(x1,x2) = 0
          comp_2!_A(x1,x2) = max{6, x1 - 19}
          comp_3#!_A(x1,x2,x3) = 5
          map_1!_A(x1) = x1 + 4
          map_2#!_A(x1,x2) = 3
          cons_2!_A(x1,x2) = 2
          app!_A(x1,x2) = max{4, x1 - 15, x2 - 22}
          comp_0!_A = 1
          comp_1!_A(x1) = max{7, x1 - 18}
          cons_1!_A(x1) = max{3, x1 - 23}
          map_2!_A(x1,x2) = max{7, x1 + 5, x2 - 22}
          nil_0!_A = 6
  
    partial status:
  
      pi(app#) = [1]
      pi(comp_2) = [1, 2]
      pi(comp_3#) = [1, 2]
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app) = []
      pi(comp_0) = []
      pi(comp_1) = []
      pi(cons_1) = [1]
      pi(map_2) = [1]
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x))
r12: twice_1(f) -> comp_2(f,f)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x))
r12: twice_1(f) -> comp_2(f,f)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{x1 + 6, x2 + 14}
        cons_2_A(x1,x2) = max{4, x1 - 8, x2 + 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{3, x1 - 2, x2 - 3}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{0, x1 - 9, x2 - 12}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.