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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev2#_A(x1,x2) = x2 + 6
          cons_A(x1,x2) = max{8, x2 + 7}
          rev#_A(x1) = max{9, x1 + 6}
          rev2_A(x1,x2) = x2
          rev_A(x1) = max{1, x1}
          nil_A = 3
          rev1_A(x1,x2) = max{5, x2 + 2}
          |0|_A = 0
          s_A(x1) = 0
          rand_A(x1) = x1 + 7
    
      precedence:
      
        rev2# = rev2 > cons = nil > rev > rev# = rev1 = |0| = s = rand
      
      partial status:
    
        pi(rev2#) = [2]
        pi(cons) = []
        pi(rev#) = [1]
        pi(rev2) = []
        pi(rev) = [1]
        pi(nil) = []
        pi(rev1) = [2]
        pi(|0|) = []
        pi(s) = []
        pi(rand) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev2#_A(x1,x2) = 8
          cons_A(x1,x2) = 11
          rev#_A(x1) = x1 + 11
          rev2_A(x1,x2) = 13
          rev_A(x1) = max{8, x1 + 2}
          nil_A = 15
          rev1_A(x1,x2) = 25
          |0|_A = 26
          s_A(x1) = 11
          rand_A(x1) = 0
    
      precedence:
      
        rev2 = nil > rev > rev2# = cons = rev# = rev1 = |0| = s > rand
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev#) = [1]
        pi(rev2) = []
        pi(rev) = []
        pi(nil) = []
        pi(rev1) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: rev#(cons(x,l)) -> rev2#(x,l)
p2: 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}


-- Reduction pair.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: 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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev2#_A(x1,x2) = max{0, x2 - 1}
          cons_A(x1,x2) = max{3, x2 + 2}
          rev_A(x1) = max{3, x1}
          nil_A = 4
          rev1_A(x1,x2) = max{6, x2 - 3}
          rev2_A(x1,x2) = x2
          |0|_A = 5
          s_A(x1) = 0
          rand_A(x1) = x1
    
      precedence:
      
        rev1 = rev2 > rev = nil > cons > rand > rev2# = |0| = s
      
      partial status:
    
        pi(rev2#) = []
        pi(cons) = []
        pi(rev) = [1]
        pi(nil) = []
        pi(rev1) = []
        pi(rev2) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev2#_A(x1,x2) = 24
          cons_A(x1,x2) = 26
          rev_A(x1) = x1 + 27
          nil_A = 11
          rev1_A(x1,x2) = 26
          rev2_A(x1,x2) = 9
          |0|_A = 25
          s_A(x1) = 25
          rand_A(x1) = 0
    
      precedence:
      
        rev2# = cons = rev = nil = rev1 = rev2 = |0| = s > rand
      
      partial status:
    
        pi(rev2#) = []
        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.

-- 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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev1#_A(x1,x2) = max{3, x2}
          cons_A(x1,x2) = x2 + 3
          rev_A(x1) = max{1, x1}
          nil_A = 5
          rev1_A(x1,x2) = max{7, x2 - 4}
          rev2_A(x1,x2) = x2
          |0|_A = 6
          s_A(x1) = 0
          rand_A(x1) = x1
    
      precedence:
      
        rev2 > rev1 > |0| > rev = nil > cons > rev1# > s = rand
      
      partial status:
    
        pi(rev1#) = [2]
        pi(cons) = []
        pi(rev) = [1]
        pi(nil) = []
        pi(rev1) = []
        pi(rev2) = []
        pi(|0|) = []
        pi(s) = []
        pi(rand) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rev1#_A(x1,x2) = max{24, x2 - 1}
          cons_A(x1,x2) = 26
          rev_A(x1) = x1 + 27
          nil_A = 11
          rev1_A(x1,x2) = 26
          rev2_A(x1,x2) = 9
          |0|_A = 25
          s_A(x1) = 25
          rand_A(x1) = 0
    
      precedence:
      
        rev1# = cons = rev = nil = rev1 = rev2 = |0| = s > rand
      
      partial status:
    
        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.