YES

We show the termination of the relative TRS R/S:

  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)))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal 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)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5}
  {p3}


-- Reduction pair.

Consider the non-minimal 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)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        rev2#_A(x1,x2) = x2 + 1
        cons_A(x1,x2) = x2 + 2
        rev#_A(x1) = x1
        rev2_A(x1,x2) = x2
        rev_A(x1) = x1
        nil_A() = 1
        rev1_A(x1,x2) = 2
        |0|_A() = 1
        s_A(x1) = 0
        rand_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > |0| > rev1 > nil > rev2 > rev > rev2# > rev# > cons
      
      argument filter:
    
        pi(rev2#) = []
        pi(cons) = 2
        pi(rev#) = []
        pi(rev2) = 2
        pi(rev) = 1
        pi(nil) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

Consider the non-minimal 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)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        rev1#_A(x1,x2) = x2
        cons_A(x1,x2) = x2 + 1
        rev_A(x1) = x1
        nil_A() = 1
        rev1_A(x1,x2) = x2 + 1
        rev2_A(x1,x2) = x2
        |0|_A() = 2
        s_A(x1) = 0
        rand_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rev2 > rev > cons > rand > rev1 > |0| > nil > rev1#
      
      argument filter:
    
        pi(rev1#) = []
        pi(cons) = []
        pi(rev) = [1]
        pi(nil) = []
        pi(rev1) = []
        pi(rev2) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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