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

  r2, r3, r4, r5, r6, r7

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3

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: 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > rev1#
      
      argument filter:
    
        pi(rev1#) = 2
        pi(cons) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > rev1#
      
      argument filter:
    
        pi(rev1#) = 2
        pi(cons) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        cons > rev1#
      
      argument filter:
    
        pi(rev1#) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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