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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,0),(1,0)) x1
        s_A(x1) = x1
        minus_A(x1,x2) = x1 + (0,1)
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,1)
        plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
        app_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,1)
        sum_A(x1) = x1 + (2,0)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = x1 + (1,2)
        minus_A(x1,x2) = x1 + (1,0)
        |0|_A() = (2,0)
        quot_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (1,0)
        plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,1),(0,1)) x2 + (1,1)
        app_A(x1,x2) = ((1,0),(1,0)) x2 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = (2,2)
        sum_A(x1) = (3,1)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = (2,1)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (0,2)
        quot_A(x1,x2) = (3,1)
        plus_A(x1,x2) = ((1,0),(1,1)) x2 + (3,1)
        app_A(x1,x2) = (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = (2,2)
        sum_A(x1) = (3,1)
        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#(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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,0)) x1
        s_A(x1) = ((0,1),(1,0)) x1
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        plus_A(x1,x2) = ((1,1),(1,1)) x2 + (1,1)
        |0|_A() = (1,0)
        quot_A(x1,x2) = ((1,1),(1,1)) x2 + (2,2)
        app_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        nil_A() = (1,0)
        cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,1)) x2 + (0,1)
        sum_A(x1) = ((0,0),(1,1)) x1
        rand_A(x1) = ((1,1),(1,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = (0,0)
        s_A(x1) = (0,1)
        minus_A(x1,x2) = (1,1)
        plus_A(x1,x2) = ((0,1),(1,1)) x2 + (0,1)
        |0|_A() = (0,2)
        quot_A(x1,x2) = ((0,0),(1,0)) x2 + (0,1)
        app_A(x1,x2) = x1 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = (1,2)
        sum_A(x1) = (2,2)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
        minus_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (2,1)
        |0|_A() = (1,2)
        quot_A(x1,x2) = (2,1)
        app_A(x1,x2) = ((1,1),(0,0)) x1 + (0,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = (2,2)
        sum_A(x1) = (1,1)
        rand_A(x1) = (0,0)
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2
        s_A(x1) = x1
        minus_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (0,1)
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,1)
        plus_A(x1,x2) = x1 + x2 + (1,1)
        app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = x1 + x2 + (2,1)
        sum_A(x1) = x1 + (0,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,0),(1,0)) x1 + x2
        s_A(x1) = x1 + (1,1)
        minus_A(x1,x2) = x1 + (1,0)
        |0|_A() = (1,1)
        quot_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        plus_A(x1,x2) = ((1,1),(1,0)) x1 + x2 + (3,1)
        app_A(x1,x2) = ((1,1),(1,0)) x1 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = ((1,1),(1,0)) x1 + ((1,1),(1,0)) x2 + (2,2)
        sum_A(x1) = ((1,1),(0,1)) x1
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + ((0,1),(0,0)) x2
        s_A(x1) = ((1,1),(0,0)) x1 + (3,3)
        minus_A(x1,x2) = ((0,1),(0,0)) x1 + (1,0)
        |0|_A() = (1,1)
        quot_A(x1,x2) = ((1,1),(0,1)) x1 + (0,1)
        plus_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,1)) x2 + (1,1)
        app_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (1,1)
        sum_A(x1) = ((0,1),(1,1)) x1 + (2,0)
        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: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = ((0,1),(0,0)) x1
        app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
        cons_A(x1,x2) = ((0,0),(1,0)) x2 + (3,0)
        sum_A(x1) = (4,1)
        minus_A(x1,x2) = x1 + x2 + (1,1)
        |0|_A() = (1,1)
        s_A(x1) = x1
        quot_A(x1,x2) = ((0,1),(1,0)) x2 + (2,1)
        plus_A(x1,x2) = x1 + x2 + (1,1)
        nil_A() = (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = (0,0)
        app_A(x1,x2) = x1 + (1,1)
        cons_A(x1,x2) = (1,1)
        sum_A(x1) = (2,1)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (1,0)
        s_A(x1) = (1,0)
        quot_A(x1,x2) = (2,1)
        plus_A(x1,x2) = x1 + x2 + (1,1)
        nil_A() = (1,1)
        rand_A(x1) = x1 + (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = (0,0)
        app_A(x1,x2) = ((0,0),(1,0)) x1 + (3,1)
        cons_A(x1,x2) = (2,2)
        sum_A(x1) = (1,1)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
        s_A(x1) = (1,1)
        quot_A(x1,x2) = (2,1)
        plus_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,0)) x2 + (0,1)
        nil_A() = (1,1)
        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: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = x1
        cons_A(x1,x2) = ((0,0),(1,1)) x1 + x2 + (1,3)
        plus_A(x1,x2) = x1 + x2 + (1,1)
        minus_A(x1,x2) = x1 + x2 + (1,1)
        |0|_A() = (1,1)
        s_A(x1) = x1
        quot_A(x1,x2) = ((0,1),(1,0)) x2 + (2,1)
        app_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (1,0)
        nil_A() = (1,0)
        sum_A(x1) = ((0,1),(0,1)) x1
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = ((0,1),(0,0)) x1
        cons_A(x1,x2) = x2 + (2,1)
        plus_A(x1,x2) = ((1,1),(1,1)) x2 + (3,1)
        minus_A(x1,x2) = (0,1)
        |0|_A() = (4,0)
        s_A(x1) = ((0,1),(0,0)) x1 + (1,1)
        quot_A(x1,x2) = (3,1)
        app_A(x1,x2) = x1 + ((0,0),(1,0)) x2
        nil_A() = (1,1)
        sum_A(x1) = (1,0)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        sum#_A(x1) = (0,0)
        cons_A(x1,x2) = (2,2)
        plus_A(x1,x2) = ((0,0),(1,1)) x2 + (2,1)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (3,2)
        s_A(x1) = (1,1)
        quot_A(x1,x2) = (2,1)
        app_A(x1,x2) = x1 + (1,1)
        nil_A() = (1,1)
        sum_A(x1) = (1,1)
        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: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,0),(1,0)) x1
        s_A(x1) = x1
        minus_A(x1,x2) = x1 + (0,1)
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,1)
        plus_A(x1,x2) = x1 + x2
        app_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (0,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
        sum_A(x1) = ((0,1),(0,1)) x1
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,0),(1,0)) x1
        s_A(x1) = x1 + (1,1)
        minus_A(x1,x2) = x1 + (0,1)
        |0|_A() = (1,3)
        quot_A(x1,x2) = ((1,0),(1,0)) x1 + (1,1)
        plus_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
        app_A(x1,x2) = x1 + (0,1)
        nil_A() = (1,1)
        cons_A(x1,x2) = x1 + (1,2)
        sum_A(x1) = (2,2)
        rand_A(x1) = (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,0)) x1
        s_A(x1) = (0,2)
        minus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
        |0|_A() = (1,1)
        quot_A(x1,x2) = x1 + (1,0)
        plus_A(x1,x2) = ((0,1),(0,0)) x1 + (1,1)
        app_A(x1,x2) = x1 + (1,1)
        nil_A() = (0,1)
        cons_A(x1,x2) = (2,2)
        sum_A(x1) = (1,2)
        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: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = ((0,1),(0,1)) x1
        cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,1)) x2 + (0,1)
        minus_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (2,1)
        |0|_A() = (1,1)
        s_A(x1) = x1
        quot_A(x1,x2) = ((1,1),(0,1)) x2 + (2,1)
        plus_A(x1,x2) = x2 + (1,1)
        app_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
        nil_A() = (1,0)
        sum_A(x1) = ((0,0),(1,1)) x1
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = (0,0)
        cons_A(x1,x2) = (0,0)
        minus_A(x1,x2) = (1,0)
        |0|_A() = (0,2)
        s_A(x1) = x1
        quot_A(x1,x2) = ((1,0),(1,0)) x2 + (1,1)
        plus_A(x1,x2) = ((0,1),(1,0)) x2 + (1,1)
        app_A(x1,x2) = ((1,1),(0,0)) x1 + (1,0)
        nil_A() = (1,1)
        sum_A(x1) = (1,1)
        rand_A(x1) = ((0,0),(1,1)) x1 + (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = (0,0)
        cons_A(x1,x2) = (0,0)
        minus_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
        s_A(x1) = ((1,0),(1,0)) x1 + (0,1)
        quot_A(x1,x2) = ((0,1),(1,1)) x2 + (1,2)
        plus_A(x1,x2) = (1,2)
        app_A(x1,x2) = (1,1)
        nil_A() = (0,1)
        sum_A(x1) = (1,1)
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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