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()))

-- SCC decomposition.

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

p1: app#(app(app(compose(),f),g),x) -> app#(g,app(f,x))
p2: app#(app(app(compose(),f),g),x) -> app#(f,x)
p3: app#(reverse(),l) -> app#(app(reverse2(),l),nil())
p4: app#(reverse(),l) -> app#(reverse2(),l)
p5: app#(app(reverse2(),app(app(cons(),x),xs)),l) -> app#(app(reverse2(),xs),app(app(cons(),x),l))
p6: app#(app(reverse2(),app(app(cons(),x),xs)),l) -> app#(reverse2(),xs)
p7: app#(app(reverse2(),app(app(cons(),x),xs)),l) -> app#(app(cons(),x),l)
p8: last#() -> app#(app(compose(),hd()),reverse())
p9: last#() -> app#(compose(),hd())
p10: init#() -> app#(app(compose(),reverse()),app(app(compose(),tl()),reverse()))
p11: init#() -> app#(compose(),reverse())
p12: init#() -> app#(app(compose(),tl()),reverse())
p13: init#() -> app#(compose(),tl())

and R consists of:

r1: app(app(app(compose(),f),g),x) -> app(g,app(f,x))
r2: app(reverse(),l) -> app(app(reverse2(),l),nil())
r3: app(app(reverse2(),nil()),l) -> l
r4: app(app(reverse2(),app(app(cons(),x),xs)),l) -> app(app(reverse2(),xs),app(app(cons(),x),l))
r5: app(hd(),app(app(cons(),x),xs)) -> x
r6: app(tl(),app(app(cons(),x),xs)) -> xs
r7: last() -> app(app(compose(),hd()),reverse())
r8: init() -> app(app(compose(),reverse()),app(app(compose(),tl()),reverse()))

The estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p5}


-- Reduction pair.

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

p1: app#(app(app(compose(),f),g),x) -> app#(g,app(f,x))
p2: app#(app(app(compose(),f),g),x) -> app#(f,x)

and R consists of:

r1: app(app(app(compose(),f),g),x) -> app(g,app(f,x))
r2: app(reverse(),l) -> app(app(reverse2(),l),nil())
r3: app(app(reverse2(),nil()),l) -> l
r4: app(app(reverse2(),app(app(cons(),x),xs)),l) -> app(app(reverse2(),xs),app(app(cons(),x),l))
r5: app(hd(),app(app(cons(),x),xs)) -> x
r6: app(tl(),app(app(cons(),x),xs)) -> xs
r7: last() -> app(app(compose(),hd()),reverse())
r8: init() -> app(app(compose(),reverse()),app(app(compose(),tl()),reverse()))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2
      app_A(x1,x2) = x1 + x2 + (1,1)
      compose_A() = (1,0)
      reverse_A() = (4,1)
      reverse2_A() = (1,0)
      nil_A() = (1,1)
      cons_A() = (1,0)
      hd_A() = (1,0)
      tl_A() = (1,1)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(app(reverse2(),app(app(cons(),x),xs)),l) -> app#(app(reverse2(),xs),app(app(cons(),x),l))

and R consists of:

r1: app(app(app(compose(),f),g),x) -> app(g,app(f,x))
r2: app(reverse(),l) -> app(app(reverse2(),l),nil())
r3: app(app(reverse2(),nil()),l) -> l
r4: app(app(reverse2(),app(app(cons(),x),xs)),l) -> app(app(reverse2(),xs),app(app(cons(),x),l))
r5: app(hd(),app(app(cons(),x),xs)) -> x
r6: app(tl(),app(app(cons(),x),xs)) -> xs
r7: last() -> app(app(compose(),hd()),reverse())
r8: init() -> app(app(compose(),reverse()),app(app(compose(),tl()),reverse()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0),(1,0)) x1 + x2
      app_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (0,1)
      reverse2_A() = (1,1)
      cons_A() = (1,0)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.