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, 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) = max{0, x1 - 5, x2 - 4}
            s_A(x1) = max{2, x1}
            minus_A(x1,x2) = max{4, x1}
            |0|_A = 2
            plus_A(x1,x2) = max{84, x1, x2}
            quot#!_A(x1,x2) = x2 + 2
            s!_A(x1) = 0
            minus!_A(x1,x2) = 0
            |0|!_A = 0
            plus!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              quot#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              minus_A(x1,x2) = x1 + 2
              |0|_A() = 2
              plus_A(x1,x2) = x1 + 2
              quot#!_A(x1,x2) = 5
              s!_A(x1) = x1 + 4
              minus!_A(x1,x2) = x1 + 2
              |0|!_A() = 0
              plus!_A(x1,x2) = x1 + 3
        
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(plus) = [1]

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

  (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:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{3, x1 - 5}
            minus_A(x1,x2) = max{x1 + 3, x2 - 4}
            plus_A(x1,x2) = max{x1, x2}
            minus#!_A(x1,x2) = 11
            s!_A(x1) = max{13, x1 - 2}
            minus!_A(x1,x2) = 12
            plus!_A(x1,x2) = 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = x1
              s_A(x1) = x1 + 2
              minus_A(x1,x2) = x1 + 1
              plus_A(x1,x2) = x1 + 3
              minus#!_A(x1,x2) = x1 + 7
              s!_A(x1) = 10
              minus!_A(x1,x2) = x1 + 9
              plus!_A(x1,x2) = 8
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []
      pi(minus) = []
      pi(plus) = []

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

  r8, r9, r10, r11, r12

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 - 22}
            app_A(x1,x2) = max{x1 + 21, x2}
            cons_A(x1,x2) = max{14, x2}
            sum_A(x1) = max{21, x1}
            nil_A = 7
            plus_A(x1,x2) = max{25, x1 - 19, x2 - 19}
            sum#!_A(x1) = max{4, x1 + 3}
            app!_A(x1,x2) = max{x1 + 25, x2 + 3}
            cons!_A(x1,x2) = x2 + 3
            sum!_A(x1) = max{17, x1 - 3}
            nil!_A = 3
            plus!_A(x1,x2) = max{52, x1 - 3, x2 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = x1
              app_A(x1,x2) = x1 + x2 + 1
              cons_A(x1,x2) = x2 + 3
              sum_A(x1) = x1
              nil_A() = 1
              plus_A(x1,x2) = x1 + 4
              sum#!_A(x1) = 9
              app!_A(x1,x2) = x1 + x2 + 6
              cons!_A(x1,x2) = x2 + 5
              sum!_A(x1) = 7
              nil!_A() = 0
              plus!_A(x1,x2) = 4
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = []

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) = 0
            cons_A(x1,x2) = max{x1 - 9, x2 + 3}
            plus_A(x1,x2) = max{x1 + 13, x2 + 2}
            |0|_A = 18
            s_A(x1) = max{4, x1 - 14}
            sum#!_A(x1) = 49
            cons!_A(x1,x2) = max{x1 + 47, x2 + 47}
            plus!_A(x1,x2) = max{36, x1 + 32, x2 + 31}
            |0|!_A = 3
            s!_A(x1) = x1 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = 0
              cons_A(x1,x2) = x2 + 6
              plus_A(x1,x2) = x1 + x2 + 4
              |0|_A() = 2
              s_A(x1) = 2
              sum#!_A(x1) = x1 + 7
              cons!_A(x1,x2) = x1 + 14
              plus!_A(x1,x2) = 13
              |0|!_A() = 11
              s!_A(x1) = 12
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [1, 2]
      pi(plus) = [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) = 0
            s_A(x1) = max{6, x1 - 7}
            plus#!_A(x1,x2) = 3
            s!_A(x1) = max{11, x1 + 2}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              s_A(x1) = x1 + 2
              plus#!_A(x1,x2) = x1 + 4
              s!_A(x1) = x1 + 5
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]

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 - 6}
            cons_A(x1,x2) = max{20, x1 - 12, x2 + 5}
            app#!_A(x1,x2) = max{2, x1 - 21}
            cons!_A(x1,x2) = max{2, x1 - 39}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              cons_A(x1,x2) = x1 + x2 + 1
              app#!_A(x1,x2) = x1 + 3
              cons!_A(x1,x2) = 3
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.