YES
We show the termination of the TRS R:

  app(app(app(compose(),f),g),x) -> app(g,app(f,x))
  app(reverse(),l) -> app(app(reverse2(),l),nil())
  app(app(reverse2(),nil()),l) -> l
  app(app(reverse2(),app(app(cons(),x),xs)),l) -> app(app(reverse2(),xs),app(app(cons(),x),l))
  app(hd(),app(app(cons(),x),xs)) -> x
  app(tl(),app(app(cons(),x),xs)) -> xs
  last() -> app(app(compose(),hd()),reverse())
  init() -> app(app(compose(),reverse()),app(app(compose(),tl()),reverse()))

Instead we prove termination of the uncurried system:

  app(compose_0(),y) -> compose_1(y)
  app(compose_1(x1),y) -> compose_2(x1,y)
  app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(hd_0(),y) -> hd_1(y)
  app(reverse_0(),y) -> reverse_1(y)
  app(reverse2_0(),y) -> reverse2_1(y)
  app(reverse2_1(x1),y) -> reverse2_2(x1,y)
  app(tl_0(),y) -> tl_1(y)
  compose_3(f,g,x) -> app(g,app(f,x))
  reverse_1(l) -> reverse2_2(l,nil_0())
  reverse2_2(nil_0(),l) -> l
  reverse2_2(cons_2(x,xs),l) -> reverse2_2(xs,cons_2(x,l))
  hd_1(cons_2(x,xs)) -> x
  tl_1(cons_2(x,xs)) -> xs
  last_0() -> compose_2(hd_0(),reverse_0())
  init_0() -> compose_2(reverse_0(),compose_2(tl_0(),reverse_0()))


-- SCC decomposition.

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

p1: app#(compose_2(x1,x2),y) -> compose_3#(x1,x2,y)
p2: app#(hd_0(),y) -> hd_1#(y)
p3: app#(reverse_0(),y) -> reverse_1#(y)
p4: app#(reverse2_1(x1),y) -> reverse2_2#(x1,y)
p5: app#(tl_0(),y) -> tl_1#(y)
p6: compose_3#(f,g,x) -> app#(g,app(f,x))
p7: compose_3#(f,g,x) -> app#(f,x)
p8: reverse_1#(l) -> reverse2_2#(l,nil_0())
p9: reverse2_2#(cons_2(x,xs),l) -> reverse2_2#(xs,cons_2(x,l))

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(hd_0(),y) -> hd_1(y)
r7: app(reverse_0(),y) -> reverse_1(y)
r8: app(reverse2_0(),y) -> reverse2_1(y)
r9: app(reverse2_1(x1),y) -> reverse2_2(x1,y)
r10: app(tl_0(),y) -> tl_1(y)
r11: compose_3(f,g,x) -> app(g,app(f,x))
r12: reverse_1(l) -> reverse2_2(l,nil_0())
r13: reverse2_2(nil_0(),l) -> l
r14: reverse2_2(cons_2(x,xs),l) -> reverse2_2(xs,cons_2(x,l))
r15: hd_1(cons_2(x,xs)) -> x
r16: tl_1(cons_2(x,xs)) -> xs
r17: last_0() -> compose_2(hd_0(),reverse_0())
r18: init_0() -> compose_2(reverse_0(),compose_2(tl_0(),reverse_0()))

The estimated dependency graph contains the following SCCs:

  {p1, p6, p7}
  {p9}


-- Reduction pair.

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

p1: app#(compose_2(x1,x2),y) -> compose_3#(x1,x2,y)
p2: compose_3#(f,g,x) -> app#(f,x)
p3: compose_3#(f,g,x) -> app#(g,app(f,x))

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(hd_0(),y) -> hd_1(y)
r7: app(reverse_0(),y) -> reverse_1(y)
r8: app(reverse2_0(),y) -> reverse2_1(y)
r9: app(reverse2_1(x1),y) -> reverse2_2(x1,y)
r10: app(tl_0(),y) -> tl_1(y)
r11: compose_3(f,g,x) -> app(g,app(f,x))
r12: reverse_1(l) -> reverse2_2(l,nil_0())
r13: reverse2_2(nil_0(),l) -> l
r14: reverse2_2(cons_2(x,xs),l) -> reverse2_2(xs,cons_2(x,l))
r15: hd_1(cons_2(x,xs)) -> x
r16: tl_1(cons_2(x,xs)) -> xs
r17: last_0() -> compose_2(hd_0(),reverse_0())
r18: init_0() -> compose_2(reverse_0(),compose_2(tl_0(),reverse_0()))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16

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
          compose_2_A(x1,x2) = x1 + x2 + 4
          compose_3#_A(x1,x2,x3) = x1 + x2 + x3 + 3
          app_A(x1,x2) = x1 + x2 + 2
          compose_0_A() = 4
          compose_1_A(x1) = x1 + 5
          compose_3_A(x1,x2,x3) = x1 + x2 + x3 + 5
          cons_0_A() = 1
          cons_1_A(x1) = x1 + 2
          cons_2_A(x1,x2) = x1 + x2 + 3
          hd_0_A() = 1
          hd_1_A(x1) = x1 + 2
          reverse_0_A() = 3
          reverse_1_A(x1) = x1 + 4
          reverse2_0_A() = 2
          reverse2_1_A(x1) = x1 + 1
          reverse2_2_A(x1,x2) = x1 + x2 + 2
          tl_0_A() = 1
          tl_1_A(x1) = x1 + 2
          nil_0_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 10, x2 - 14}
          compose_2_A(x1,x2) = max{9, x1, x2}
          compose_3#_A(x1,x2,x3) = max{0, x1 - 10, x2 - 10, x3 - 14}
          app_A(x1,x2) = max{13, x1 + 3, x2}
          compose_0_A = 0
          compose_1_A(x1) = max{4, x1}
          compose_3_A(x1,x2,x3) = max{13, x1 + 3, x2 + 3, x3}
          cons_0_A = 1
          cons_1_A(x1) = max{5, x1}
          cons_2_A(x1,x2) = max{9, x1 + 3, x2}
          hd_0_A = 1
          hd_1_A(x1) = max{5, x1}
          reverse_0_A = 5
          reverse_1_A(x1) = max{13, x1}
          reverse2_0_A = 0
          reverse2_1_A(x1) = max{4, x1}
          reverse2_2_A(x1,x2) = max{8, x1, x2}
          tl_0_A = 1
          tl_1_A(x1) = max{5, x1}
          nil_0_A = 10
          app#!_A(x1,x2) = max{0, x1 - 10, x2 - 18}
          compose_2!_A(x1,x2) = max{3, x1 - 2, x2 - 6}
          compose_3#!_A(x1,x2,x3) = max{4, x1 - 14, x2 - 10}
          app!_A(x1,x2) = max{0, x1 - 13, x2 - 14}
          compose_0!_A = 1
          compose_1!_A(x1) = max{2, x1 - 3}
          compose_3!_A(x1,x2,x3) = max{0, x2 - 5}
          cons_0!_A = 1
          cons_1!_A(x1) = max{2, x1 - 12}
          cons_2!_A(x1,x2) = max{3, x1 - 11, x2 - 22}
          hd_0!_A = 1
          hd_1!_A(x1) = max{0, x1 - 14}
          reverse_0!_A = 1
          reverse_1!_A(x1) = max{2, x1 - 12}
          reverse2_0!_A = 1
          reverse2_1!_A(x1) = max{0, x1 - 14}
          reverse2_2!_A(x1,x2) = max{0, x1 - 13, x2 - 21}
          tl_0!_A = 1
          tl_1!_A(x1) = max{0, x1 - 14}
          nil_0!_A = 3
  
    partial status:
  
      pi(app#) = [2]
      pi(compose_2) = [1, 2]
      pi(compose_3#) = []
      pi(app) = [1]
      pi(compose_0) = []
      pi(compose_1) = [1]
      pi(compose_3) = [3]
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(cons_2) = [2]
      pi(hd_0) = []
      pi(hd_1) = []
      pi(reverse_0) = []
      pi(reverse_1) = [1]
      pi(reverse2_0) = []
      pi(reverse2_1) = []
      pi(reverse2_2) = []
      pi(tl_0) = []
      pi(tl_1) = []
      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: reverse2_2#(cons_2(x,xs),l) -> reverse2_2#(xs,cons_2(x,l))

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(hd_0(),y) -> hd_1(y)
r7: app(reverse_0(),y) -> reverse_1(y)
r8: app(reverse2_0(),y) -> reverse2_1(y)
r9: app(reverse2_1(x1),y) -> reverse2_2(x1,y)
r10: app(tl_0(),y) -> tl_1(y)
r11: compose_3(f,g,x) -> app(g,app(f,x))
r12: reverse_1(l) -> reverse2_2(l,nil_0())
r13: reverse2_2(nil_0(),l) -> l
r14: reverse2_2(cons_2(x,xs),l) -> reverse2_2(xs,cons_2(x,l))
r15: hd_1(cons_2(x,xs)) -> x
r16: tl_1(cons_2(x,xs)) -> xs
r17: last_0() -> compose_2(hd_0(),reverse_0())
r18: init_0() -> compose_2(reverse_0(),compose_2(tl_0(),reverse_0()))

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:
          reverse2_2#_A(x1,x2) = x1 + x2 + 1
          cons_2_A(x1,x2) = x1 + x2 + 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          reverse2_2#_A(x1,x2) = max{0, x1 - 9}
          cons_2_A(x1,x2) = max{10, x1 + 8, x2 + 8}
          reverse2_2#!_A(x1,x2) = max{6, x1 - 1}
          cons_2!_A(x1,x2) = 5
  
    partial status:
  
      pi(reverse2_2#) = []
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.