YES

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

  R:
  p(s(x)) -> x
  fac(|0|()) -> s(|0|())
  fac(s(x)) -> times(s(x),fac(p(s(x))))

  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: fac#(s(x)) -> fac#(p(s(x)))
p2: fac#(s(x)) -> p#(s(x))

and R consists of:

r1: p(s(x)) -> x
r2: fac(|0|()) -> s(|0|())
r3: fac(s(x)) -> times(s(x),fac(p(s(x))))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: fac#(s(x)) -> fac#(p(s(x)))

and R consists of:

r1: p(s(x)) -> x
r2: fac(|0|()) -> s(|0|())
r3: fac(s(x)) -> times(s(x),fac(p(s(x))))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = x1
          s_A(x1) = x1
          p_A(x1) = x1
          fac_A(x1) = x1 + 1
          |0|_A = 1
          times_A(x1,x2) = max{x1 + 1, x2}
          rand_A(x1) = x1 + 2
    
      precedence:
      
        fac# = s = p = fac = |0| = times = rand
      
      partial status:
    
        pi(fac#) = []
        pi(s) = []
        pi(p) = []
        pi(fac) = []
        pi(|0|) = []
        pi(times) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = max{0, x1 - 8}
          s_A(x1) = x1 + 10
          p_A(x1) = max{5, x1 - 10}
          fac_A(x1) = max{11, x1 + 7}
          |0|_A = 2
          times_A(x1,x2) = 17
          rand_A(x1) = 1
    
      precedence:
      
        p > s = fac = |0| = times > fac# = rand
      
      partial status:
    
        pi(fac#) = []
        pi(s) = []
        pi(p) = []
        pi(fac) = [1]
        pi(|0|) = []
        pi(times) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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