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, r4, r5, r8, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{3, x1}
        compose_2_A(x1,x2) = max{x1 + 6, x2 + 5}
        compose_3#_A(x1,x2,x3) = max{x1 + 6, x2 + 4}
        app_A(x1,x2) = max{3, x1 - 7, x2 - 7}
        compose_0_A = 1
        compose_1_A(x1) = max{2, x1 - 7}
        cons_0_A = 6
        cons_1_A(x1) = max{1, x1 - 8}
        cons_2_A(x1,x2) = max{2, x1 - 16, x2 - 17}
        reverse2_0_A = 1
        reverse2_1_A(x1) = max{2, x1 - 8}
        last_0_A = 8
        hd_0_A = 1
        reverse_0_A = 1
        init_0_A = 13
        tl_0_A = 1
  
    base order 2:
  
      precedence:
      
        last_0 = init_0 > app# = compose_2 > compose_3# = app = tl_0 > compose_1 = cons_0 = cons_2 = reverse2_1 = reverse_0 > compose_0 = cons_1 = reverse2_0 = hd_0
      
  
    partial status:
  
      pi(app#) = [1]
      pi(compose_2) = [2]
      pi(compose_3#) = [2]
      pi(app) = []
      pi(compose_0) = []
      pi(compose_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(reverse2_0) = []
      pi(reverse2_1) = []
      pi(last_0) = []
      pi(hd_0) = []
      pi(reverse_0) = []
      pi(init_0) = []
      pi(tl_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: compose_3#(f,g,x) -> app#(f,x)
p2: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  r1, r2, r8, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        reverse2_2#_A(x1,x2) = max{x1 + 14, x2 + 11}
        cons_2_A(x1,x2) = max{12, x1 - 12, x2}
        app_A(x1,x2) = max{15, x1 - 4, x2 - 17}
        compose_0_A = 1
        compose_1_A(x1) = max{2, x1 - 18}
        compose_2_A(x1,x2) = max{3, x1 - 22, x2 - 17}
        reverse2_0_A = 2
        reverse2_1_A(x1) = max{3, x1 - 18}
        last_0_A = 4
        hd_0_A = 3
        reverse_0_A = 3
        init_0_A = 4
        tl_0_A = 1
  
    base order 2:
  
      precedence:
      
        app > reverse2_2# = last_0 = init_0 > cons_2 = compose_2 = reverse_0 = tl_0 > compose_1 = reverse2_1 = hd_0 > compose_0 = reverse2_0
      
  
    partial status:
  
      pi(reverse2_2#) = [1]
      pi(cons_2) = [2]
      pi(app) = []
      pi(compose_0) = []
      pi(compose_1) = []
      pi(compose_2) = []
      pi(reverse2_0) = []
      pi(reverse2_1) = []
      pi(last_0) = []
      pi(hd_0) = []
      pi(reverse_0) = []
      pi(init_0) = []
      pi(tl_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.