YES

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

  R:
  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(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#(x,s(y)) -> pred#(minus(x,y))
p2: minus#(x,s(y)) -> minus#(x,y)
p3: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p4: quot#(s(x),s(y)) -> minus#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p3}
  {p2}


-- 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: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = ((1,1),(0,0)) x1
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
        pred_A(x1) = ((1,0),(1,0)) x1
        |0|_A() = (1,0)
        quot_A(x1,x2) = x1 + (1,0)
        rand_A(x1) = ((1,1),(0,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = x1 + (1,1)
        minus_A(x1,x2) = x1
        pred_A(x1) = x1
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + (1,1)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,1),(0,0)) x1
        s_A(x1) = ((0,1),(0,0)) x1 + (2,1)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        pred_A(x1) = x1
        |0|_A() = (1,2)
        quot_A(x1,x2) = x1 + (2,1)
        rand_A(x1) = (0,0)
    

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#(x,s(y)) -> minus#(x,y)

and R consists of:

r1: pred(s(x)) -> x
r2: minus(x,|0|()) -> x
r3: minus(x,s(y)) -> pred(minus(x,y))
r4: quot(|0|(),s(y)) -> |0|()
r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,0),(1,1)) x2
        s_A(x1) = x1
        pred_A(x1) = x1
        minus_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (0,1)
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,0)) x2
        s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        pred_A(x1) = x1
        minus_A(x1,x2) = x1
        |0|_A() = (1,1)
        quot_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
        rand_A(x1) = x1 + (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((0,1),(0,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        pred_A(x1) = x1
        minus_A(x1,x2) = x1
        |0|_A() = (0,1)
        quot_A(x1,x2) = ((0,1),(1,0)) x1
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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