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))
  sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
  pred(cons(s(x),nil())) -> cons(x,nil())


-- 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)
p8: sum#(plus(cons(|0|(),x),cons(y,l))) -> pred#(sum(cons(s(x),cons(y,l))))
p9: sum#(plus(cons(|0|(),x),cons(y,l))) -> sum#(cons(s(x),cons(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))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())

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))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())

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) = max{0, x1 - 108}
            app_A(x1,x2) = max{x1 + 107, x2 + 34}
            cons_A(x1,x2) = max{39, x1 + 3, x2}
            sum_A(x1) = max{72, x1}
            nil_A = 1
            plus_A(x1,x2) = max{69, x1 - 11, x2}
            |0|_A = 6
            s_A(x1) = max{69, x1 - 70}
            sum#!_A(x1) = max{1, x1 - 25}
            app!_A(x1,x2) = max{x1 + 157, x2 + 85}
            cons!_A(x1,x2) = x2 + 29
            sum!_A(x1) = 31
            nil!_A = 3
            plus!_A(x1,x2) = max{1, x1 - 8}
            |0|!_A = 1
            s!_A(x1) = 30
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 10}
            app_A(x1,x2) = max{2, x1 + 1, x2 + 1}
            cons_A(x1,x2) = max{x1, x2}
            sum_A(x1) = max{8, x1}
            nil_A = 2
            plus_A(x1,x2) = max{8, x1 - 10, x2}
            |0|_A = 0
            s_A(x1) = max{2, x1 - 10}
            sum#!_A(x1) = 6
            app!_A(x1,x2) = max{x1 + 12, x2 + 4}
            cons!_A(x1,x2) = 5
            sum!_A(x1) = 7
            nil!_A = 3
            plus!_A(x1,x2) = max{8, x1 - 1}
            |0|!_A = 1
            s!_A(x1) = x1 + 9
        
  
    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))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())

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 - 12}
            cons_A(x1,x2) = max{40, x1 - 56, x2 + 11}
            app#!_A(x1,x2) = max{1, x1 - 9}
            cons!_A(x1,x2) = max{16, x1 - 99, x2 + 2}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 1}
            cons_A(x1,x2) = max{x1 - 20, x2}
            app#!_A(x1,x2) = max{4, x1 - 1, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 69, x2 - 48}
        
  
    partial status:
  
      pi(app#) = []
      pi(cons) = []

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))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())

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 - 32}
            cons_A(x1,x2) = max{22, x1 - 4, x2 + 2}
            plus_A(x1,x2) = max{35, x1 - 62, x2 + 1}
            |0|_A = 7
            s_A(x1) = max{26, x1 - 36}
            sum#!_A(x1) = max{1, x1 - 29}
            cons!_A(x1,x2) = max{4, x1 - 65, x2 - 17}
            plus!_A(x1,x2) = max{0, x2 - 22}
            |0|!_A = 0
            s!_A(x1) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 13}
            cons_A(x1,x2) = max{7, x1 - 5, x2 + 5}
            plus_A(x1,x2) = max{6, x1 - 18, x2 + 2}
            |0|_A = 0
            s_A(x1) = max{3, x1 - 8}
            sum#!_A(x1) = max{1, x1 - 13}
            cons!_A(x1,x2) = max{0, x1 - 37}
            plus!_A(x1,x2) = max{2, x1 - 4}
            |0|!_A = 1
            s!_A(x1) = 4
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = []
      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))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())

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 - 2}
            s_A(x1) = max{40, x1 + 1}
            plus#!_A(x1,x2) = max{17, x1 - 41, x2 - 1}
            s!_A(x1) = max{16, x1 - 43}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 2}
            s_A(x1) = max{78, x1 + 1}
            plus#!_A(x1,x2) = max{24, x1 - 79, x2 - 1}
            s!_A(x1) = max{23, x1 - 95}
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.