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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev2#_A(x1,x2) = 1
            cons_A(x1,x2) = max{14, x1 - 4, x2}
            rev#_A(x1) = 1
            rev2_A(x1,x2) = 14
            rev_A(x1) = 14
            nil_A = 14
            rev1_A(x1,x2) = 16
            |0|_A = 16
            s_A(x1) = 16
            rev2#!_A(x1,x2) = max{x1 + 8, x2 + 12}
            cons!_A(x1,x2) = max{x1 + 5, x2 + 20}
            rev#!_A(x1) = x1 + 12
            rev2!_A(x1,x2) = x1 + 36
            rev!_A(x1) = 35
            nil!_A = 2
            rev1!_A(x1,x2) = 5
            |0|!_A = 1
            s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev2#_A(x1,x2) = 1
            cons_A(x1,x2) = max{18, x2 + 15}
            rev#_A(x1) = 1
            rev2_A(x1,x2) = max{3, x2}
            rev_A(x1) = max{16, x1}
            nil_A = 6
            rev1_A(x1,x2) = 0
            |0|_A = 0
            s_A(x1) = 0
            rev2#!_A(x1,x2) = max{9, x2 - 4}
            cons!_A(x1,x2) = x1 + 6
            rev#!_A(x1) = max{10, x1 - 13}
            rev2!_A(x1,x2) = 8
            rev!_A(x1) = 7
            nil!_A = 7
            rev1!_A(x1,x2) = 5
            |0|!_A = 6
            s!_A(x1) = 1
        
  
    partial status:
  
      pi(rev2#) = []
      pi(cons) = []
      pi(rev#) = []
      pi(rev2) = [1, 2]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev1#_A(x1,x2) = 0
            cons_A(x1,x2) = max{3, x1 - 17, x2 - 2}
            rev1#!_A(x1,x2) = max{x1 + 2, x2 + 19}
            cons!_A(x1,x2) = max{40, x1 + 31, x2 + 29}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev1#_A(x1,x2) = 0
            cons_A(x1,x2) = max{7, x1 - 14}
            rev1#!_A(x1,x2) = max{1, x2 - 12}
            cons!_A(x1,x2) = max{2, x2 - 1}
        
  
    partial status:
  
      pi(rev1#) = [2]
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.