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, 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) = x2 + 7
            rev#_A(x1) = 1
            rev2_A(x1,x2) = x2
            rev_A(x1) = max{1, x1}
            nil_A = 10
            rev1_A(x1,x2) = max{13, x1 + 1, x2 - 9}
            rev2#!_A(x1,x2) = max{2, x2 - 3}
            cons!_A(x1,x2) = max{3, x2 - 1}
            rev#!_A(x1) = max{1, x1 - 6}
            rev2!_A(x1,x2) = max{14, x2 + 5}
            rev!_A(x1) = 13
            nil!_A = 15
            rev1!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              rev2#_A(x1,x2) = 0
              cons_A(x1,x2) = 4
              rev#_A(x1) = 0
              rev2_A(x1,x2) = x1 + x2 + 1
              rev_A(x1) = 4
              nil_A() = 0
              rev1_A(x1,x2) = x2
              rev2#!_A(x1,x2) = x1 + x2 + 5
              cons!_A(x1,x2) = 6
              rev#!_A(x1) = x1
              rev2!_A(x1,x2) = 8
              rev!_A(x1) = 7
              nil!_A() = 1
              rev1!_A(x1,x2) = 1
        
  
    partial status:
  
      pi(rev2#) = []
      pi(cons) = []
      pi(rev#) = [1]
      pi(rev2) = [1]
      pi(rev) = [1]
      pi(nil) = []
      pi(rev1) = []

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{15, x1 - 5, x2 - 5}
            rev1#!_A(x1,x2) = 6
            cons!_A(x1,x2) = max{x1 + 5, x2 + 5}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              rev1#_A(x1,x2) = x2
              cons_A(x1,x2) = x1 + x2 + 1
              rev1#!_A(x1,x2) = x2 + 1
              cons!_A(x1,x2) = x2
        
  
    partial status:
  
      pi(rev1#) = []
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.