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

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        rev2#_A(x1,x2) = max{14, x2 + 7}
        cons_A(x1,x2) = max{13, x2 + 8}
        rev#_A(x1) = max{14, x1 + 5}
        rev2_A(x1,x2) = max{5, x2}
        rev_A(x1) = max{11, x1}
        nil_A = 10
        rev1_A(x1,x2) = 12
        |0|_A = 9
        s_A(x1) = 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          rev2#_A(x1,x2) = max{0, x1 - 15, x2 - 10}
          cons_A(x1,x2) = max{9, x1 - 5, x2}
          rev#_A(x1) = max{0, x1 - 10}
          rev2_A(x1,x2) = max{9, x1 - 15, x2 - 11}
          rev_A(x1) = max{9, x1 - 10}
          nil_A = 9
          rev1_A(x1,x2) = max{3, x2 - 5}
          |0|_A = 4
          s_A(x1) = 4
          rev2#!_A(x1,x2) = max{15, x1 + 3, x2 + 6}
          cons!_A(x1,x2) = x1 + 3
          rev#!_A(x1) = 16
          rev2!_A(x1,x2) = max{16, x1 + 5, x2 + 10}
          rev!_A(x1) = x1 + 10
          nil!_A = 3
          rev1!_A(x1,x2) = max{11, x2 - 9}
          |0|!_A = 1
          s!_A(x1) = max{1, x1 - 1}
  
    partial status:
  
      pi(rev2#) = []
      pi(cons) = []
      pi(rev#) = []
      pi(rev2) = []
      pi(rev) = []
      pi(nil) = []
      pi(rev1) = []
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1, p2, p3

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        rev1#_A(x1,x2) = max{x1 - 4, x2 + 6}
        cons_A(x1,x2) = max{x1 - 4, x2 + 7}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          rev1#_A(x1,x2) = 0
          cons_A(x1,x2) = max{5, x1 - 10, x2 - 4}
          rev1#!_A(x1,x2) = max{1, x1 - 13, x2 - 9}
          cons!_A(x1,x2) = max{3, x1 - 12, x2 - 8}
  
    partial status:
  
      pi(rev1#) = [2]
      pi(cons) = []

The following rules are removed:

  p1

Then no dependency pair remains.