YES

We show the termination of the TRS R:

  rev1(|0|(),nil()) -> |0|()
  rev1(s(X),nil()) -> s(X)
  rev1(X,cons(Y,L)) -> rev1(Y,L)
  rev(nil()) -> nil()
  rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
  rev2(X,nil()) -> nil()
  rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

-- SCC decomposition.

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

p1: rev1#(X,cons(Y,L)) -> rev1#(Y,L)
p2: rev#(cons(X,L)) -> rev1#(X,L)
p3: rev#(cons(X,L)) -> rev2#(X,L)
p4: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))
p5: rev2#(X,cons(Y,L)) -> rev#(rev2(Y,L))
p6: rev2#(X,cons(Y,L)) -> rev2#(Y,L)

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p6}
  {p1}


-- Reduction pair.

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

p1: rev2#(X,cons(Y,L)) -> rev#(rev2(Y,L))
p2: rev#(cons(X,L)) -> rev2#(X,L)
p3: rev2#(X,cons(Y,L)) -> rev2#(Y,L)
p4: rev2#(X,cons(Y,L)) -> rev#(cons(X,rev(rev2(Y,L))))

and R consists of:

r1: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    rev2#_A(x1,x2) = x2 + 2
    cons_A(x1,x2) = max{6, x2 + 5}
    rev#_A(x1) = x1
    rev2_A(x1,x2) = x2
    rev_A(x1) = max{1, x1}
    rev1_A(x1,x2) = max{0, x2 - 9}
    |0|_A = 0
    nil_A = 6
    s_A(x1) = 0

The next rules are strictly ordered:

  p1, p2, p3, p4

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: rev1(|0|(),nil()) -> |0|()
r2: rev1(s(X),nil()) -> s(X)
r3: rev1(X,cons(Y,L)) -> rev1(Y,L)
r4: rev(nil()) -> nil()
r5: rev(cons(X,L)) -> cons(rev1(X,L),rev2(X,L))
r6: rev2(X,nil()) -> nil()
r7: rev2(X,cons(Y,L)) -> rev(cons(X,rev(rev2(Y,L))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    rev1#_A(x1,x2) = max{x1 + 1, x2}
    cons_A(x1,x2) = max{x1 + 2, x2 + 1}

The next rules are strictly ordered:

  p1

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