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

  r5

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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 32}
            comp_2_A(x1,x2) = max{x1 + 28, x2 + 3}
            comp_3#_A(x1,x2,x3) = max{1, x1 - 4, x2 - 29}
            map_1_A(x1) = x1 + 31
            map_2#_A(x1,x2) = max{1, x1 - 32}
            cons_2_A(x1,x2) = max{8, x1 - 9}
            app_A(x1,x2) = max{6, x1 - 32}
            cons_1_A(x1) = max{40, x1 + 23}
            app#!_A(x1,x2) = max{25, x1 - 4}
            comp_2!_A(x1,x2) = max{x1 - 2, x2 + 25}
            comp_3#!_A(x1,x2,x3) = max{25, x1 - 3, x2 - 1}
            map_1!_A(x1) = 4
            map_2#!_A(x1,x2) = x1 + 26
            cons_2!_A(x1,x2) = 17
            app!_A(x1,x2) = max{3, x1 - 24, x2 - 1}
            cons_1!_A(x1) = max{17, x1 + 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = 0
              comp_2_A(x1,x2) = 4
              comp_3#_A(x1,x2,x3) = 0
              map_1_A(x1) = 3
              map_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 2
              app_A(x1,x2) = 2
              cons_1_A(x1) = 3
              app#!_A(x1,x2) = 2
              comp_2!_A(x1,x2) = 4
              comp_3#!_A(x1,x2,x3) = 2
              map_1!_A(x1) = 1
              map_2#!_A(x1,x2) = 0
              cons_2!_A(x1,x2) = 4
              app!_A(x1,x2) = 3
              cons_1!_A(x1) = x1 + 5
        
  
    partial status:
  
      pi(app#) = []
      pi(comp_2) = [2]
      pi(comp_3#) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app) = [1, 2]
      pi(cons_1) = [1]

The following rules are removed:

  p3, p5



-- 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: comp_3#(f,g,x) -> app#(g,x)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p4}
  {p3}


-- 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#(f,app(g,x))
p3: 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 set of usable rules consists of

  r9

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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 1
            comp_2_A(x1,x2) = max{x1 + 6, x2 + 19}
            comp_3#_A(x1,x2,x3) = 1
            app_A(x1,x2) = max{0, x2 - 8}
            map_2_A(x1,x2) = max{3, x2 - 10}
            nil_0_A = 1
            app#!_A(x1,x2) = max{5, x1 - 4}
            comp_2!_A(x1,x2) = 2
            comp_3#!_A(x1,x2,x3) = max{x1 + 1, x2 + 6}
            app!_A(x1,x2) = max{12, x1 - 5, x2 + 6}
            map_2!_A(x1,x2) = x1 + 11
            nil_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = 0
              comp_2_A(x1,x2) = x2
              comp_3#_A(x1,x2,x3) = 0
              app_A(x1,x2) = x2
              map_2_A(x1,x2) = 2
              nil_0_A() = 1
              app#!_A(x1,x2) = 1
              comp_2!_A(x1,x2) = 1
              comp_3#!_A(x1,x2,x3) = x1 + x2 + x3 + 2
              app!_A(x1,x2) = x2 + 4
              map_2!_A(x1,x2) = x1 + x2 + 1
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(comp_2) = []
      pi(comp_3#) = []
      pi(app) = [2]
      pi(map_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{7, x1 - 7, x2 - 7}
            map_2#!_A(x1,x2) = 5
            cons_2!_A(x1,x2) = max{4, x1 - 6}
        
        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) = x2 + 2
              map_2#!_A(x1,x2) = x2 + 7
              cons_2!_A(x1,x2) = 6
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []

The following rules are removed:

  p1

Then no dependency pair remains.