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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 1, x2 + 1}
        s_A(x1) = x1
        times_A(x1,x2) = max{x1 + 3, x2 + 2}
        |0|_A = 1
        plus_A(x1,x2) = max{x1, x2 + 3}
        rand_A(x1) = x1 + 2
  
    precedence:
    
      times > plus > s > times# = |0| = rand
    
    partial status:
  
      pi(times#) = [2]
      pi(s) = [1]
      pi(times) = [2]
      pi(|0|) = []
      pi(plus) = [1, 2]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 1, x2 + 1}
        s_A(x1) = x1
        times_A(x1,x2) = max{x1 + 3, x2 + 3}
        |0|_A = 1
        plus_A(x1,x2) = max{x1, x2 + 3}
        rand_A(x1) = x1 + 2
  
    precedence:
    
      times = rand > plus > plus# > s > |0|
    
    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

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  {p1}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        plus#_A(x1,x2) = x1
        s_A(x1) = x1
        times_A(x1,x2) = max{x1 + 2, x2 + 1}
        |0|_A = 1
        plus_A(x1,x2) = max{2, x1, x2}
        rand_A(x1) = x1 + 2
  
    precedence:
    
      times > plus > s > |0| > plus# = rand
    
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(times) = [2]
      pi(|0|) = []
      pi(plus) = [1, 2]
      pi(rand) = []

The next rules are strictly ordered:

  p1

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