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:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{1, x1}
        app_A(x1,x2) = max{x1 + 3, x2 + 4}
        cons_A(x1,x2) = x2
        sum_A(x1) = max{0, x1 - 8}
        nil_A = 0
        plus_A(x1,x2) = max{12, x1 - 12, x2 + 9}
        |0|_A = 0
        s_A(x1) = max{10, x1 - 13}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = max{0, x1 - 30}
          app_A(x1,x2) = max{16, x1 + 1, x2 + 13}
          cons_A(x1,x2) = max{14, x1 - 14, x2}
          sum_A(x1) = max{16, x1}
          nil_A = 1
          plus_A(x1,x2) = max{12, x1 - 4, x2}
          |0|_A = 0
          s_A(x1) = max{1, x1 - 13}
          sum#!_A(x1) = x1 + 11
          app!_A(x1,x2) = max{69, x1 + 13, x2 + 53}
          cons!_A(x1,x2) = max{0, x2 - 6}
          sum!_A(x1) = 0
          nil!_A = 1
          plus!_A(x1,x2) = max{2, x1 - 11, x2 - 20}
          |0|!_A = 0
          s!_A(x1) = max{1, x1 - 21}
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{1, x1 - 1}
        cons_A(x1,x2) = max{9, x2 + 3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 3}
          cons_A(x1,x2) = max{x1 + 4, x2 + 2}
          app#!_A(x1,x2) = max{18, x1 - 5, x2 - 1}
          cons!_A(x1,x2) = max{x1 + 11, x2 + 17}
  
    partial status:
  
      pi(app#) = []
      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:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = x1 + 4
        cons_A(x1,x2) = max{x1 + 6, x2 + 16}
        plus_A(x1,x2) = max{21, x1 - 30, x2 + 7}
        |0|_A = 0
        s_A(x1) = max{19, x1 - 31}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = max{0, x1 - 25}
          cons_A(x1,x2) = max{x1 - 95, x2 + 12}
          plus_A(x1,x2) = max{107, x1 - 122, x2 + 12}
          |0|_A = 0
          s_A(x1) = max{13, x1 - 108}
          sum#!_A(x1) = 4
          cons!_A(x1,x2) = max{3, x1 - 229, x2 - 6}
          plus!_A(x1,x2) = max{5, x1 - 230, x2 - 340}
          |0|!_A = 0
          s!_A(x1) = max{4, x1 - 446}
  
    partial status:
  
      pi(sum#) = [1]
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{0, x2 - 1}
        s_A(x1) = max{6, x1 - 43}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 11}
          s_A(x1) = max{68, x1 + 10}
          plus#!_A(x1,x2) = max{1, x1 - 66}
          s!_A(x1) = max{0, x1 - 109}
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.