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)))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  minus(minus(x,y),z) -> minus(x,plus(y,z))
  app(nil(),k) -> k
  app(l,nil()) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil())) -> cons(x,nil())
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))

  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)
p4: plus#(s(x),y) -> plus#(x,y)
p5: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p6: minus#(minus(x,y),z) -> plus#(y,z)
p7: app#(cons(x,l),k) -> app#(l,k)
p8: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p9: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p10: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p11: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p12: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}
  {p10}
  {p8}
  {p4}
  {p7}


-- 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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          quot#_A(x1,x2) = x1 + (0,1)
          s_A(x1) = x1 + (0,1)
          minus_A(x1,x2) = x1
          |0|_A() = (1,2)
          quot_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,2)
          plus_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (3,5)
          app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,1)
          nil_A() = (1,5)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
          sum_A(x1) = x1 + (0,1)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,2)
  
    precedence:
    
      minus = |0| = quot = plus = app = nil = cons = sum = rand > quot# = s
    
    partial status:
  
      pi(quot#) = [1]
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []
      pi(plus) = []
      pi(app) = []
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      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)
p2: minus#(minus(x,y),z) -> minus#(x,plus(y,z))

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          minus#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,6)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          minus_A(x1,x2) = ((1,0),(0,0)) x1 + (3,5)
          plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (7,3)
          |0|_A() = (1,0)
          quot_A(x1,x2) = ((0,0),(1,0)) x2 + (2,4)
          app_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,3)
          nil_A() = (1,0)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (1,2)
          sum_A(x1) = (5,1)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      minus = plus > |0| = app > quot > minus# = s = nil = sum > cons = rand
    
    partial status:
  
      pi(minus#) = []
      pi(s) = []
      pi(minus) = []
      pi(plus) = []
      pi(|0|) = []
      pi(quot) = []
      pi(app) = [1]
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      pi(rand) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: 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: 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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + (1,3)
          s_A(x1) = x1 + (0,2)
          minus_A(x1,x2) = x1
          |0|_A() = (0,0)
          quot_A(x1,x2) = x1 + (1,1)
          plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,0)
          app_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,0)
          nil_A() = (5,1)
          cons_A(x1,x2) = (1,0)
          sum_A(x1) = (4,0)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,3)
  
    precedence:
    
      s = minus = |0| = quot = plus = app = nil = rand > minus# = sum > cons
    
    partial status:
  
      pi(minus#) = [2]
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []
      pi(plus) = []
      pi(app) = []
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      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: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = x1 + (0,1)
          app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,6)
          cons_A(x1,x2) = ((0,0),(1,0)) x2 + (4,7)
          sum_A(x1) = (4,9)
          minus_A(x1,x2) = x1 + (1,1)
          |0|_A() = (0,0)
          s_A(x1) = x1
          quot_A(x1,x2) = (0,0)
          plus_A(x1,x2) = ((1,0),(1,1)) x2 + (3,3)
          nil_A() = (1,10)
          rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      app = sum = plus = nil = rand > sum# > cons = minus = |0| = s = quot
    
    partial status:
  
      pi(sum#) = [1]
      pi(app) = []
      pi(cons) = []
      pi(sum) = []
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(plus) = []
      pi(nil) = []
      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: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = x1 + (1,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (3,1)
          plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,11)
          minus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
          |0|_A() = (0,0)
          s_A(x1) = ((1,0),(0,0)) x1
          quot_A(x1,x2) = (0,1)
          app_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,3)
          nil_A() = (1,0)
          sum_A(x1) = ((1,0),(0,0)) x1 + (0,10)
          rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      plus = quot = app = sum > cons > sum# = minus = |0| = s = nil = rand
    
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = []
      pi(plus) = []
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(app) = []
      pi(nil) = []
      pi(sum) = []
      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#(s(x),y) -> plus#(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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x1 + (0,1)
          s_A(x1) = x1 + (0,2)
          minus_A(x1,x2) = x1
          |0|_A() = (0,0)
          quot_A(x1,x2) = x1 + (1,1)
          plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,3)
          app_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,3)
          nil_A() = (1,1)
          cons_A(x1,x2) = (1,0)
          sum_A(x1) = ((1,0),(0,0)) x1 + (0,2)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
  
    precedence:
    
      minus = |0| = plus = sum = rand > plus# = nil > quot > s = cons > app
    
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = [1]
      pi(plus) = []
      pi(app) = [1, 2]
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      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: app#(cons(x,l),k) -> app#(l,k)

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: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: rand(x) -> x
r15: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + (1,1)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (2,7)
          minus_A(x1,x2) = ((1,0),(0,0)) x1 + (3,29)
          |0|_A() = (1,0)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          quot_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (2,1)
          plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (7,30)
          app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (3,0)
          nil_A() = (1,1)
          sum_A(x1) = x1 + (0,15)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
  
    precedence:
    
      minus = |0| = quot = app = rand > s = plus > nil = sum > cons > app#
    
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(plus) = [2]
      pi(app) = []
      pi(nil) = []
      pi(sum) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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