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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            average#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,3)
            s_A(x1) = x1 + (0,2)
            average_A(x1,x2) = x1 + x2 + (1,0)
            |0|_A() = (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        |0| = rand > average > average# = s
      
      partial status:
    
        pi(average#) = [1, 2]
        pi(s) = [1]
        pi(average) = []
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            average#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,0)
            average_A(x1,x2) = (4,0)
            |0|_A() = (1,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        |0| = rand > average# = s = average
      
      partial status:
    
        pi(average#) = [2]
        pi(s) = []
        pi(average) = []
        pi(|0|) = []
        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: 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}


-- Reduction pair.

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

p1: 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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            average#_A(x1,x2) = x2
            s_A(x1) = x1 + (0,2)
            average_A(x1,x2) = x1 + x2 + (1,1)
            |0|_A() = (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        average# = s = average = |0| = rand
      
      partial status:
    
        pi(average#) = []
        pi(s) = []
        pi(average) = []
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            average#_A(x1,x2) = ((1,0),(0,0)) x2
            s_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            average_A(x1,x2) = ((1,0),(0,0)) x1 + (1,3)
            |0|_A() = (1,1)
            rand_A(x1) = (0,0)
    
      precedence:
      
        rand > average# = s = average > |0|
      
      partial status:
    
        pi(average#) = []
        pi(s) = []
        pi(average) = []
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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