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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = x2
            s_A(x1) = max{5, x1}
            minus_A(x1,x2) = max{x1 + 5, x2 - 6}
            |0|_A = 2
            plus_A(x1,x2) = max{x1, x2}
            quot#!_A(x1,x2) = 0
            s!_A(x1) = 3
            minus!_A(x1,x2) = 0
            |0|!_A = 5
            plus!_A(x1,x2) = max{29, x1 - 4, x2 + 2}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 8, x2 - 6}
            s_A(x1) = max{4, x1}
            minus_A(x1,x2) = max{7, x1, x2 - 46}
            |0|_A = 8
            plus_A(x1,x2) = max{4, x1 - 56, x2}
            quot#!_A(x1,x2) = 0
            s!_A(x1) = max{1, x1 - 10}
            minus!_A(x1,x2) = max{2, x1 - 19, x2 - 9}
            |0|!_A = 2
            plus!_A(x1,x2) = max{52, x1 - 5, x2 - 5}
        
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(plus) = [1, 2]

The following rules are removed:

  p1

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

  r1, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 6, x2 - 4}
            s_A(x1) = max{2, x1}
            minus_A(x1,x2) = max{4, x1 + 1, x2 - 4}
            plus_A(x1,x2) = max{3, x1 - 6, x2}
            |0|_A = 2
            minus#!_A(x1,x2) = max{3, x1 - 6, x2 - 4}
            s!_A(x1) = max{2, x1 - 18}
            minus!_A(x1,x2) = max{5, x1 - 7, x2 - 13}
            plus!_A(x1,x2) = max{4, x1 - 12, x2 - 14}
            |0|!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 25, x2 - 32}
            s_A(x1) = max{21, x1}
            minus_A(x1,x2) = max{23, x1 + 1, x2 - 32}
            plus_A(x1,x2) = max{31, x1 - 33, x2}
            |0|_A = 3
            minus#!_A(x1,x2) = max{1, x1 - 28, x2 - 35}
            s!_A(x1) = max{2, x1 - 71}
            minus!_A(x1,x2) = max{4, x1 - 29, x2 - 69}
            plus!_A(x1,x2) = max{3, x1 - 62, x2 - 70}
            |0|!_A = 2
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(minus) = [1, 2]
      pi(plus) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p1, p2

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, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{62, x1}
            app_A(x1,x2) = max{x1 + 70, x2}
            cons_A(x1,x2) = 62
            sum_A(x1) = max{2, x1}
            plus_A(x1,x2) = max{25, x1 - 70, x2 + 12}
            |0|_A = 13
            s_A(x1) = max{13, x1 - 26}
            nil_A = 61
            sum#!_A(x1) = 0
            app!_A(x1,x2) = x2
            cons!_A(x1,x2) = 0
            sum!_A(x1) = max{0, x1 - 62}
            plus!_A(x1,x2) = max{x1 - 14, x2 + 55}
            |0|!_A = 30
            s!_A(x1) = x1 + 29
            nil!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{14, x1 - 30}
            app_A(x1,x2) = max{x1 + 29, x2 + 16}
            cons_A(x1,x2) = max{29, x1 + 13, x2}
            sum_A(x1) = max{13, x1}
            plus_A(x1,x2) = max{14, x2}
            |0|_A = 5
            s_A(x1) = max{4, x1 - 15}
            nil_A = 1
            sum#!_A(x1) = x1 + 3
            app!_A(x1,x2) = max{63, x1 + 34, x2 + 31}
            cons!_A(x1,x2) = x2 + 4
            sum!_A(x1) = 6
            plus!_A(x1,x2) = max{3, x1 - 14, x2 - 10}
            |0|!_A = 1
            s!_A(x1) = max{0, x1 - 23}
            nil!_A = 1
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 15}
            cons_A(x1,x2) = max{x1 - 13, x2 + 3}
            plus_A(x1,x2) = max{27, x1 - 43, x2 + 2}
            |0|_A = 3
            s_A(x1) = max{3, x1 - 46}
            sum#!_A(x1) = max{30, x1 - 7}
            cons!_A(x1,x2) = max{x1 + 4, x2 + 45}
            plus!_A(x1,x2) = max{3, x1 - 51, x2 - 52}
            |0|!_A = 3
            s!_A(x1) = max{1, x1 - 54}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 23}
            cons_A(x1,x2) = max{x1 - 60, x2 + 8}
            plus_A(x1,x2) = max{82, x1 - 72, x2 + 5}
            |0|_A = 92
            s_A(x1) = max{6, x1 - 83}
            sum#!_A(x1) = max{3, x1 - 276}
            cons!_A(x1,x2) = max{3, x1 - 332, x2 - 270}
            plus!_A(x1,x2) = max{4, x1 - 165, x2 - 250}
            |0|!_A = 2
            s!_A(x1) = max{1, x1 - 331}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [1, 2]
      pi(plus) = [1, 2]
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 6}
            s_A(x1) = max{16, x1 + 5}
            plus#!_A(x1,x2) = max{2, x1 - 6}
            s!_A(x1) = max{1, x1 - 76}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 4}
            s_A(x1) = max{17, x1 + 3}
            plus#!_A(x1,x2) = max{1, x1 - 78, x2 - 1}
            s!_A(x1) = max{0, x1 - 78}
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 5}
            cons_A(x1,x2) = max{x1 - 40, x2 + 4}
            app#!_A(x1,x2) = max{4, x1 - 296, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 339, x2 - 297}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 5}
            cons_A(x1,x2) = max{24, x1 - 12, x2 + 4}
            app#!_A(x1,x2) = max{4, x1 - 134, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 239, x2 - 133}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.