YES

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

  R:
  times(x,|0|()) -> |0|()
  times(x,s(y)) -> plus(times(x,y),x)
  plus(x,|0|()) -> x
  plus(|0|(),x) -> x
  plus(x,s(y)) -> s(plus(x,y))
  plus(s(x),y) -> s(plus(x,y))

  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: times#(x,s(y)) -> plus#(times(x,y),x)
p2: times#(x,s(y)) -> times#(x,y)
p3: plus#(x,s(y)) -> plus#(x,y)
p4: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3, p4}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = max{4, x2 + 1}
          s_A(x1) = max{3, x1}
          times_A(x1,x2) = max{x1 + 2, x2 + 4}
          |0|_A = 1
          plus_A(x1,x2) = max{x1, x2 + 2}
          rand_A(x1) = max{8, x1 + 4}
    
      precedence:
      
        times# = times > |0| = plus > s = rand
      
      partial status:
    
        pi(times#) = [2]
        pi(s) = [1]
        pi(times) = [1, 2]
        pi(|0|) = []
        pi(plus) = [1, 2]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = max{5, x2 + 1}
          s_A(x1) = max{4, x1 + 3}
          times_A(x1,x2) = max{x1 + 2, x2 + 1}
          |0|_A = 1
          plus_A(x1,x2) = x2 + 3
          rand_A(x1) = 1
    
      precedence:
      
        plus > times# = s = times = |0| = rand
      
      partial status:
    
        pi(times#) = [2]
        pi(s) = [1]
        pi(times) = [2]
        pi(|0|) = []
        pi(plus) = []
        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: plus#(x,s(y)) -> plus#(x,y)
p2: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{x1 + 4, x2 + 1}
          s_A(x1) = max{2, x1}
          times_A(x1,x2) = max{x1 + 4, x2 + 2}
          |0|_A = 5
          plus_A(x1,x2) = max{x1, x2 + 2}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        times > |0| = plus > s > plus# > rand
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
        pi(times) = [1, 2]
        pi(|0|) = []
        pi(plus) = [1, 2]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{5, x1, x2}
          s_A(x1) = x1 + 5
          times_A(x1,x2) = max{x1 + 8, x2 + 3}
          |0|_A = 7
          plus_A(x1,x2) = max{8, x1, x2 + 3}
          rand_A(x1) = 0
    
      precedence:
      
        plus# = s = |0| > times > plus = rand
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
        pi(times) = [2]
        pi(|0|) = []
        pi(plus) = [1, 2]
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2

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