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:

  lexicographic path order with precedence:
  
    precedence:
    
      nil > s > |0| > rev1 > rev2 > rev > cons > rev2# > rev#
    
    argument filter:
  
      pi(rev2#) = [2]
      pi(cons) = [2]
      pi(rev#) = [1]
      pi(rev2) = 2
      pi(rev) = 1
      pi(rev1) = 2
      pi(|0|) = []
      pi(nil) = []
      pi(s) = []

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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      cons > rev1#
    
    argument filter:
  
      pi(rev1#) = [1, 2]
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7

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