YES

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

  R:
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(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: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,1)
            s_A(x1) = x1 + (0,1)
            minus_A(x1,x2) = x1
            |0|_A() = (1,2)
            quot_A(x1,x2) = x1 + x2 + (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        quot# = s = minus = |0| = quot = rand
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(quot) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = x1 + ((1,0),(0,0)) x2
            s_A(x1) = (2,2)
            minus_A(x1,x2) = ((1,0),(1,0)) x1 + (3,0)
            |0|_A() = (0,1)
            quot_A(x1,x2) = ((0,0),(1,0)) x1 + (4,2)
            rand_A(x1) = (3,3)
    
      precedence:
      
        minus > quot# = quot > s = rand > |0|
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(quot) = []
        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: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = x1 + x2 + (0,1)
            s_A(x1) = x1 + (0,1)
            minus_A(x1,x2) = x1
            |0|_A() = (1,0)
            quot_A(x1,x2) = x1 + (1,0)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        minus# = s = minus = |0| = quot = rand
      
      partial status:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(quot) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (1,2)
            s_A(x1) = ((1,0),(1,0)) x1 + (2,1)
            minus_A(x1,x2) = x1 + (0,6)
            |0|_A() = (0,0)
            quot_A(x1,x2) = ((1,0),(1,0)) x1 + (3,3)
            rand_A(x1) = (3,2)
    
      precedence:
      
        rand > s > minus > minus# > |0| = quot
      
      partial status:
    
        pi(minus#) = [1]
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(quot) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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