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, r6, 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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            sum#_A(0) = 0
            sum#_A(1) = 0
            app_A(0,0) = 0
            app_A(0,1) = 0
            app_A(1,0) = 0
            app_A(1,1) = 1
            cons_A(0,0) = 0
            cons_A(0,1) = 0
            cons_A(1,0) = 0
            cons_A(1,1) = 0
            sum_A(0) = 0
            sum_A(1) = 1
            nil_A = 1
            plus_A(0,0) = 0
            plus_A(0,1) = 1
            plus_A(1,0) = 1
            plus_A(1,1) = 0
            |0|_A = 0
            s_A(0) = 0
            s_A(1) = 1
            sum#!_A(0) = 0
            sum#!_A(1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 0
            cons!_A(0,0) = 1
            cons!_A(0,1) = 0
            cons!_A(1,0) = 1
            cons!_A(1,1) = 0
            sum!_A(0) = 0
            sum!_A(1) = 0
            nil!_A = 1
            plus!_A(0,0) = 1
            plus!_A(0,1) = 1
            plus!_A(1,0) = 1
            plus!_A(1,1) = 1
            |0|!_A = 1
            s!_A(0) = 1
            s!_A(1) = 1
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 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

  r1, r2, r3, r4, r5, r6, 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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            app#_A(0,0) = 0
            app#_A(0,1) = 0
            app#_A(1,0) = 0
            app#_A(1,1) = 0
            cons_A(0,0) = 1
            cons_A(0,1) = 0
            cons_A(1,0) = 1
            cons_A(1,1) = 0
            app_A(0,0) = 0
            app_A(0,1) = 1
            app_A(1,0) = 1
            app_A(1,1) = 0
            nil_A = 0
            sum_A(0) = 1
            sum_A(1) = 1
            plus_A(0,0) = 0
            plus_A(0,1) = 1
            plus_A(1,0) = 0
            plus_A(1,1) = 1
            |0|_A = 1
            s_A(0) = 0
            s_A(1) = 1
            app#!_A(0,0) = 0
            app#!_A(0,1) = 0
            app#!_A(1,0) = 0
            app#!_A(1,1) = 0
            cons!_A(0,0) = 0
            cons!_A(0,1) = 0
            cons!_A(1,0) = 0
            cons!_A(1,1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            nil!_A = 0
            sum!_A(0) = 1
            sum!_A(1) = 1
            plus!_A(0,0) = 0
            plus!_A(0,1) = 0
            plus!_A(1,0) = 0
            plus!_A(1,1) = 0
            |0|!_A = 1
            s!_A(0) = 0
            s!_A(1) = 0
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(sum) = []
      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: 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

  r1, r2, r3, r4, r5, r6, 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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            sum#_A(0) = 0
            sum#_A(1) = 0
            cons_A(0,0) = 0
            cons_A(0,1) = 0
            cons_A(1,0) = 0
            cons_A(1,1) = 0
            plus_A(0,0) = 1
            plus_A(0,1) = 0
            plus_A(1,0) = 0
            plus_A(1,1) = 1
            app_A(0,0) = 0
            app_A(0,1) = 0
            app_A(1,0) = 0
            app_A(1,1) = 1
            nil_A = 1
            sum_A(0) = 0
            sum_A(1) = 0
            |0|_A = 1
            s_A(0) = 1
            s_A(1) = 0
            sum#!_A(0) = 0
            sum#!_A(1) = 0
            cons!_A(0,0) = 0
            cons!_A(0,1) = 0
            cons!_A(1,0) = 0
            cons!_A(1,1) = 0
            plus!_A(0,0) = 0
            plus!_A(0,1) = 0
            plus!_A(1,0) = 0
            plus!_A(1,1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            nil!_A = 0
            sum!_A(0) = 1
            sum!_A(1) = 0
            |0|!_A = 1
            s!_A(0) = 0
            s!_A(1) = 0
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(plus) = [2]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(sum) = []
      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

  r1, r2, r3, r4, r5, r6, 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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            plus#_A(0,0) = 0
            plus#_A(0,1) = 0
            plus#_A(1,0) = 0
            plus#_A(1,1) = 0
            s_A(0) = 1
            s_A(1) = 1
            app_A(0,0) = 0
            app_A(0,1) = 1
            app_A(1,0) = 1
            app_A(1,1) = 1
            nil_A = 0
            cons_A(0,0) = 1
            cons_A(0,1) = 1
            cons_A(1,0) = 1
            cons_A(1,1) = 1
            sum_A(0) = 1
            sum_A(1) = 1
            plus_A(0,0) = 0
            plus_A(0,1) = 1
            plus_A(1,0) = 1
            plus_A(1,1) = 1
            |0|_A = 0
            plus#!_A(0,0) = 0
            plus#!_A(0,1) = 0
            plus#!_A(1,0) = 0
            plus#!_A(1,1) = 0
            s!_A(0) = 0
            s!_A(1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 0
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            nil!_A = 1
            cons!_A(0,0) = 0
            cons!_A(0,1) = 1
            cons!_A(1,0) = 0
            cons!_A(1,1) = 1
            sum!_A(0) = 1
            sum!_A(1) = 0
            plus!_A(0,0) = 0
            plus!_A(0,1) = 0
            plus!_A(1,0) = 1
            plus!_A(1,1) = 1
            |0|!_A = 1
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      pi(plus) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.