YES

We show the termination of the TRS 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)))))

-- SCC decomposition.

Consider the 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)))))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  r1, r2, r5, r6, r7

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) = ((0,0),(1,0)) x1 + (1,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (3,2)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (2,5)
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,1)
            |0|_A() = (1,1)
    
      precedence:
      
        s = minus = plus = |0| > quot#
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(plus) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = (0,0)
            s_A(x1) = (0,0)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
            |0|_A() = (0,0)
    
      precedence:
      
        minus > quot# = s = plus = |0|
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(plus) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{x1 + 6, x2 + 2}
          s_A(x1) = max{3, x1}
          minus_A(x1,x2) = max{x1, x2}
          plus_A(x1,x2) = max{x1 + 4, x2}
          |0|_A = 0
    
      precedence:
      
        plus > s > minus > minus# = |0|
      
      partial status:
    
        pi(minus#) = [1, 2]
        pi(s) = [1]
        pi(minus) = [1, 2]
        pi(plus) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{x1 + 5, x2 + 2}
          s_A(x1) = max{6, x1 + 3}
          minus_A(x1,x2) = max{x1 - 1, x2 + 1}
          plus_A(x1,x2) = max{x1 - 4, x2 + 3}
          |0|_A = 0
    
      precedence:
      
        s > plus > minus# = minus = |0|
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = []
        pi(minus) = [2]
        pi(plus) = [2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  r5, r6, r8, r9, r10, r11, r12

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

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = x1 + 9
          cons_A(x1,x2) = x2 + 2
          plus_A(x1,x2) = max{x1 + 8, x2 + 8}
          |0|_A = 0
          s_A(x1) = max{7, x1}
    
      precedence:
      
        sum# = cons = plus > |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = 19
          cons_A(x1,x2) = max{11, x2 + 7}
          plus_A(x1,x2) = max{12, x1 - 2, x2 + 1}
          |0|_A = 0
          s_A(x1) = max{13, x1 + 1}
    
      precedence:
      
        sum# = cons = plus = |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(plus) = []
        pi(|0|) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{2, x1 + 1, x2}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{x1 - 1, x2 + 1}
          s_A(x1) = x1
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the 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)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x1 + 1
          cons_A(x1,x2) = max{x1, x2 + 1}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = [1]
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 1}
          cons_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = []
        pi(cons) = [1, 2]
    

The next rules are strictly ordered:

  p1

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