YES
We show the termination of the TRS R:

  app(app(append(),nil()),l) -> l
  app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t))
  app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3))
  app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2))

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(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)
  append_2(nil_0(),l) -> l
  append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t))
  append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3))
  map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2))


-- 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#(map_1(x1),y) -> map_2#(x1,y)
p3: append_2#(cons_2(h,t),l) -> append_2#(t,l)
p4: map_2#(f,cons_2(h,t)) -> app#(f,h)
p5: map_2#(f,cons_2(h,t)) -> map_2#(f,t)
p6: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3))
p7: append_2#(append_2(l1,l2),l3) -> append_2#(l2,l3)
p8: map_2#(f,append_2(l1,l2)) -> append_2#(map_2(f,l1),map_2(f,l2))
p9: map_2#(f,append_2(l1,l2)) -> map_2#(f,l1)
p10: map_2#(f,append_2(l1,l2)) -> map_2#(f,l2)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: append_2(nil_0(),l) -> l
r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l))
r9: map_2(f,nil_0()) -> nil_0()
r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t))
r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3))
r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5, p9, p10}
  {p3, p6, p7}


-- Reduction pair.

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

p1: map_2#(f,cons_2(h,t)) -> app#(f,h)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,append_2(l1,l2)) -> map_2#(f,l2)
p4: map_2#(f,append_2(l1,l2)) -> map_2#(f,l1)
p5: map_2#(f,cons_2(h,t)) -> map_2#(f,t)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: append_2(nil_0(),l) -> l
r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l))
r9: map_2(f,nil_0()) -> nil_0()
r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t))
r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3))
r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2))

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) = max{0, x1 - 4, x2 - 18}
            cons_2_A(x1,x2) = max{17, x1 + 4, x2 + 4}
            app#_A(x1,x2) = max{0, x1 - 4, x2 - 18}
            map_1_A(x1) = max{3, x1 + 2}
            append_2_A(x1,x2) = max{5, x1 + 4, x2 + 4}
            map_2#!_A(x1,x2) = 3
            cons_2!_A(x1,x2) = max{x1 + 11, x2 + 5}
            app#!_A(x1,x2) = 3
            map_1!_A(x1) = max{2, x1 - 9}
            append_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:
              map_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x1 + x2
              map_1_A(x1) = x1 + 3
              append_2_A(x1,x2) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x2
              cons_2!_A(x1,x2) = 10
              app#!_A(x1,x2) = x2 + 1
              map_1!_A(x1) = 1
              append_2!_A(x1,x2) = x1 + x2 + 3
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = [1]
      pi(app#) = []
      pi(map_1) = [1]
      pi(append_2) = [1]

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: append_2#(cons_2(h,t),l) -> append_2#(t,l)
p2: append_2#(append_2(l1,l2),l3) -> append_2#(l2,l3)
p3: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3))

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: append_2(nil_0(),l) -> l
r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l))
r9: map_2(f,nil_0()) -> nil_0()
r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t))
r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3))
r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2))

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:
          
            append_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{x1, x2}
            append_2_A(x1,x2) = max{0, x2 - 6}
            append_2#!_A(x1,x2) = max{2, x1 - 1}
            cons_2!_A(x1,x2) = x2 + 10
            append_2!_A(x1,x2) = max{11, x1 + 7, x2 + 5}
        
        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) = 0
              cons_2_A(x1,x2) = x2 + 5
              append_2_A(x1,x2) = x1 + 5
              append_2#!_A(x1,x2) = 4
              cons_2!_A(x1,x2) = 5
              append_2!_A(x1,x2) = x2 + 6
        
  
    partial status:
  
      pi(append_2#) = [1]
      pi(cons_2) = [1, 2]
      pi(append_2) = [1, 2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.