YES
We show the termination of the TRS R:

  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)))))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))


-- SCC decomposition.

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

p1: app#(cons(x,l),k) -> app#(l,k)
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p3: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p4: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p5: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p6: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p7: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p7}


-- 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: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r8

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
            app_A(x1,x2) = max{x1, x2}
            cons_A(x1,x2) = max{6, x2}
            sum_A(x1) = 15
            nil_A = 3
            plus_A(x1,x2) = max{5, x1 - 7, x2}
            |0|_A = 0
            s_A(x1) = max{1, x1 - 6}
            sum#!_A(x1) = x1 + 13
            app!_A(x1,x2) = max{x1 + 29, x2 + 6}
            cons!_A(x1,x2) = max{2, x2 + 1}
            sum!_A(x1) = 5
            nil!_A = 3
            plus!_A(x1,x2) = max{11, x1 + 7, x2 + 10}
            |0|!_A = 0
            s!_A(x1) = max{9, x1 + 6}
        
        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 + 3
              cons_A(x1,x2) = x1 + x2 + 3
              sum_A(x1) = x1
              nil_A() = 3
              plus_A(x1,x2) = x2 + 1
              |0|_A() = 1
              s_A(x1) = 1
              sum#!_A(x1) = 7
              app!_A(x1,x2) = x1 + 8
              cons!_A(x1,x2) = 5
              sum!_A(x1) = x1
              nil!_A() = 0
              plus!_A(x1,x2) = x1 + 4
              |0|!_A() = 0
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      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: app#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

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 - 11}
            cons_A(x1,x2) = max{x1 - 8, x2 + 10}
            app#!_A(x1,x2) = max{7, x1 - 11, x2 - 1}
            cons!_A(x1,x2) = max{6, x1 - 27}
        
        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
              cons!_A(x1,x2) = x2 + 4
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [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: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r7, r8

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 - 179}
            cons_A(x1,x2) = max{x1 + 125, x2 + 53}
            plus_A(x1,x2) = max{36, x2 + 17}
            |0|_A = 0
            s_A(x1) = max{18, x1 - 18}
            sum#!_A(x1) = max{2, x1 - 122}
            cons!_A(x1,x2) = max{x1 + 19, x2 - 121}
            plus!_A(x1,x2) = max{0, x1 - 19}
            |0|!_A = 0
            s!_A(x1) = 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
              cons_A(x1,x2) = x1 + x2 + 2
              plus_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 1
              s_A(x1) = 1
              sum#!_A(x1) = 3
              cons!_A(x1,x2) = 2
              plus!_A(x1,x2) = x2 + 1
              |0|!_A() = 0
              s!_A(x1) = 0
        
  
    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: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

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{1, x1 - 83}
            plus#!_A(x1,x2) = max{0, x1 - 84}
            s!_A(x1) = max{1, x1 - 83}
        
        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 + 1
              plus#!_A(x1,x2) = x1
              s!_A(x1) = x1 + 4
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.