YES

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

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

  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: average#(s(x),y) -> average#(x,s(y))
p2: average#(x,s(s(s(y)))) -> average#(s(x),y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: average#(s(x),y) -> average#(x,s(y))
p2: average#(x,s(s(s(y)))) -> average#(s(x),y)

and R consists of:

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

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        rand > average > |0| > s > average#
      
      argument filter:
    
        pi(average#) = [1, 2]
        pi(s) = 1
        pi(average) = [1, 2]
        pi(|0|) = []
        pi(rand) = [1]
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        average#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,0,1)) x1 + ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x2
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,0,0)) x1 + (3,4,5,6)
        average_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,0,0,8)
        |0|_A() = (1,1,1,1)
        rand_A(x1) = (1,1,1,1)
    

The next rules are strictly ordered:

  p1, p2

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