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

  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
            s_A(x1) = x1 + (0,1)
            minus_A(x1,x2) = x1
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,7)
            |0|_A() = (1,2)
            quot_A(x1,x2) = x1 + (1,2)
            app_A(x1,x2) = x1 + x2 + (3,9)
            nil_A() = (7,10)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (2,8)
            sum_A(x1) = ((1,0),(1,0)) x1
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        sum > |0| = quot = nil = cons = rand > plus > minus# = s = minus = app
      
      partial status:
    
        pi(minus#) = []
        pi(s) = [1]
        pi(minus) = [1]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(quot) = []
        pi(app) = [1, 2]
        pi(nil) = []
        pi(cons) = []
        pi(sum) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = (0,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (3,1)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (7,2)
            |0|_A() = (1,1)
            quot_A(x1,x2) = (2,4)
            app_A(x1,x2) = x1 + (3,1)
            nil_A() = (1,0)
            cons_A(x1,x2) = (2,0)
            sum_A(x1) = (0,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        minus# = minus = plus = nil = rand > quot > cons > s = |0| = app = sum
      
      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:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 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#(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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (6,4)
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,0)
            |0|_A() = (1,5)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            quot_A(x1,x2) = ((0,0),(1,0)) x1 + (7,3)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,2)
            nil_A() = (1,3)
            cons_A(x1,x2) = (2,1)
            sum_A(x1) = (3,2)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        quot = nil = sum = rand > plus = |0| = s = app = cons > minus# = minus
      
      partial status:
    
        pi(minus#) = []
        pi(minus) = []
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(app) = [2]
        pi(nil) = []
        pi(cons) = []
        pi(sum) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = (0,0)
            minus_A(x1,x2) = (0,0)
            plus_A(x1,x2) = ((1,0),(0,0)) x2 + (3,3)
            |0|_A() = (1,1)
            s_A(x1) = (1,2)
            quot_A(x1,x2) = (2,3)
            app_A(x1,x2) = x2 + (3,2)
            nil_A() = (3,2)
            cons_A(x1,x2) = (2,0)
            sum_A(x1) = (1,1)
            rand_A(x1) = (2,3)
    
      precedence:
      
        plus = quot > app = nil > minus# = minus = |0| = s = cons = sum = rand
      
      partial status:
    
        pi(minus#) = []
        pi(minus) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(app) = [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: 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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = x1 + (1,2)
            app_A(x1,x2) = x1 + x2 + (7,4)
            cons_A(x1,x2) = x2 + (6,3)
            sum_A(x1) = (8,5)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (3,2)
            |0|_A() = (1,2)
            s_A(x1) = ((1,0),(0,0)) x1
            quot_A(x1,x2) = (2,1)
            plus_A(x1,x2) = ((1,0),(1,1)) x2 + (7,1)
            nil_A() = (1,1)
            rand_A(x1) = x1
    
      precedence:
      
        sum# = app = cons = sum = minus = |0| = s = quot = plus = nil = rand
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(plus) = []
        pi(nil) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = (0,0)
            app_A(x1,x2) = x1 + (4,4)
            cons_A(x1,x2) = (2,2)
            sum_A(x1) = (3,3)
            minus_A(x1,x2) = (1,1)
            |0|_A() = (1,1)
            s_A(x1) = (0,0)
            quot_A(x1,x2) = (0,0)
            plus_A(x1,x2) = (0,0)
            nil_A() = (1,1)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        sum = |0| = nil > minus = rand > sum# = app = cons = s = quot = plus
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [1]
        pi(cons) = []
        pi(sum) = []
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(plus) = []
        pi(nil) = []
        pi(rand) = [1]
    

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. 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)) x2 + (6,8)
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,10)
            minus_A(x1,x2) = x1
            |0|_A() = (1,2)
            s_A(x1) = x1 + (0,1)
            quot_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,0)
            app_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,7)
            nil_A() = (3,9)
            sum_A(x1) = ((0,0),(1,0)) x1 + (10,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        quot > |0| = rand > app > nil = sum > cons > sum# > plus = minus > s
      
      partial status:
    
        pi(sum#) = [1]
        pi(cons) = []
        pi(plus) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(quot) = []
        pi(app) = []
        pi(nil) = []
        pi(sum) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = (0,0)
            cons_A(x1,x2) = (2,0)
            plus_A(x1,x2) = ((1,0),(1,0)) x1 + (1,0)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + (2,1)
            |0|_A() = (1,1)
            s_A(x1) = ((0,0),(1,0)) x1 + (4,2)
            quot_A(x1,x2) = (5,0)
            app_A(x1,x2) = (4,1)
            nil_A() = (1,1)
            sum_A(x1) = (3,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        rand > plus > sum > cons = app > quot > s > minus > sum# = |0| = nil
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = []
        pi(plus) = []
        pi(minus) = [1]
        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:

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