YES

We show the termination of the TRS R:

  rev(nil()) -> nil()
  rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
  rev1(|0|(),nil()) -> |0|()
  rev1(s(x),nil()) -> s(x)
  rev1(x,cons(y,l)) -> rev1(y,l)
  rev2(x,nil()) -> nil()
  rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

-- SCC decomposition.

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

p1: rev#(cons(x,l)) -> rev1#(x,l)
p2: rev#(cons(x,l)) -> rev2#(x,l)
p3: rev1#(x,cons(y,l)) -> rev1#(y,l)
p4: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
p5: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5}
  {p3}


-- Reduction pair.

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

p1: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
p2: rev#(cons(x,l)) -> rev2#(x,l)
p3: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The set of usable rules consists of

  r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      rev2#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + (0,0,1,0)
      cons_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (2,1,1,1)
      rev#_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + (0,0,0,1)
      rev2_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (0,5,1,2)
      rev1_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,1,0)) x2 + (2,1,1,0)
      |0|_A() = (1,3,1,3)
      nil_A() = (1,1,1,1)
      s_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,3,4,3)
      rev_A(x1) = x1 + (0,3,6,2)

The next rules are strictly ordered:

  p1, p2, p3

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: rev1#(x,cons(y,l)) -> rev1#(y,l)

and R consists of:

r1: rev(nil()) -> nil()
r2: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r3: rev1(|0|(),nil()) -> |0|()
r4: rev1(s(x),nil()) -> s(x)
r5: rev1(x,cons(y,l)) -> rev1(y,l)
r6: rev2(x,nil()) -> nil()
r7: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      rev1#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,1)) x2
      cons_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)

The next rules are strictly ordered:

  p1

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