YES

We show the termination of the relative TRS R/S:

  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))

  S:
  cons(x,cons(y,l)) -> cons(y,cons(x,l))

-- SCC decomposition.

Consider the non-minimal 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))
r9: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p6}
  {p1}
  {p7}


-- Reduction pair.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p2: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p3: 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = max{5, x1 + 4}
          app_A(x1,x2) = max{72, x1 + 29, x2 + 12}
          cons_A(x1,x2) = 4
          sum_A(x1) = 32
          plus_A(x1,x2) = max{x1 + 28, x2 + 3}
          nil_A = 3
          |0|_A = 0
          s_A(x1) = max{29, x1}
    
      precedence:
      
        app > sum > cons > plus = nil = s > sum# = |0|
      
      partial status:
    
        pi(sum#) = [1]
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = [2]
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = max{6, x1 - 13}
          app_A(x1,x2) = 0
          cons_A(x1,x2) = 8
          sum_A(x1) = 9
          plus_A(x1,x2) = 78
          nil_A = 0
          |0|_A = 0
          s_A(x1) = 5
    
      precedence:
      
        sum# = app = cons = sum = plus = nil = |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = []
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimal 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)))))
p2: 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the non-minimal 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)))))
p2: 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            sum#_A(x1) = ((1,0),(0,0)) x1 + (2,14)
            app_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (4,3)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,2)
            sum_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            plus_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,0)
            nil_A() = (1,1)
            |0|_A() = (1,1)
            s_A(x1) = (1,1)
    
      precedence:
      
        sum# = cons = sum > app = nil = |0| > plus = s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [2]
        pi(cons) = []
        pi(sum) = []
        pi(plus) = [2]
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            sum#_A(x1) = (0,1)
            app_A(x1,x2) = ((1,0),(1,1)) x2 + (1,3)
            cons_A(x1,x2) = (0,2)
            sum_A(x1) = (0,6)
            plus_A(x1,x2) = ((1,1),(1,1)) x2 + (2,7)
            nil_A() = (0,0)
            |0|_A() = (1,1)
            s_A(x1) = (1,1)
    
      precedence:
      
        plus > cons = sum = nil > sum# = app = |0| > s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [2]
        pi(cons) = []
        pi(sum) = []
        pi(plus) = []
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the non-minimal 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the non-minimal 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            sum#_A(x1) = x1 + (1,1)
            app_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (1,2)
            cons_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
            sum_A(x1) = x1
            nil_A() = (1,1)
            plus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (0,3)
            |0|_A() = (1,1)
            s_A(x1) = (1,1)
    
      precedence:
      
        sum# = app = cons = sum = nil = plus = |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(nil) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            sum#_A(x1) = ((0,1),(0,0)) x1 + (1,16)
            app_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (4,4)
            cons_A(x1,x2) = ((0,1),(1,0)) x2 + (3,8)
            sum_A(x1) = (5,17)
            nil_A() = (1,1)
            plus_A(x1,x2) = (4,12)
            |0|_A() = (1,1)
            s_A(x1) = (4,12)
    
      precedence:
      
        app = sum > plus = |0| = s > cons > nil > sum#
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [1]
        pi(cons) = []
        pi(sum) = []
        pi(nil) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the non-minimal 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            app#_A(x1,x2) = ((0,1),(0,0)) x1 + (1,0)
            cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(0,1)) x2 + (2,1)
            app_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (3,0)
            nil_A() = (1,1)
            sum_A(x1) = x1
            plus_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,3)
            |0|_A() = (1,1)
            s_A(x1) = (2,1)
    
      precedence:
      
        |0| > app = sum > plus = s > nil > app# = cons
      
      partial status:
    
        pi(app#) = []
        pi(cons) = []
        pi(app) = [1, 2]
        pi(nil) = []
        pi(sum) = []
        pi(plus) = [2]
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            app#_A(x1,x2) = (1,1)
            cons_A(x1,x2) = (3,1)
            app_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2 + (4,2)
            nil_A() = (1,0)
            sum_A(x1) = (4,1)
            plus_A(x1,x2) = ((1,0),(1,1)) x2 + (2,2)
            |0|_A() = (1,1)
            s_A(x1) = (1,1)
    
      precedence:
      
        plus > |0| = s > cons = app > sum > app# = nil
      
      partial status:
    
        pi(app#) = []
        pi(cons) = []
        pi(app) = [1, 2]
        pi(nil) = []
        pi(sum) = []
        pi(plus) = [2]
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the non-minimal 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: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{1, x1}
          s_A(x1) = max{1, x1}
          app_A(x1,x2) = max{x1 + 9, x2 + 5}
          nil_A = 2
          cons_A(x1,x2) = max{3, x2}
          sum_A(x1) = max{4, x1}
          plus_A(x1,x2) = max{x1 + 10, x2 + 10}
          |0|_A = 0
    
      precedence:
      
        plus > plus# = s = nil > app = cons = sum = |0|
      
      partial status:
    
        pi(plus#) = [1]
        pi(s) = [1]
        pi(app) = []
        pi(nil) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 2}
          s_A(x1) = max{21, x1 + 1}
          app_A(x1,x2) = 4
          nil_A = 16
          cons_A(x1,x2) = 4
          sum_A(x1) = 1
          plus_A(x1,x2) = x1 + 19
          |0|_A = 0
    
      precedence:
      
        nil > plus# = s = app = cons = sum = plus = |0|
      
      partial status:
    
        pi(plus#) = []
        pi(s) = []
        pi(app) = []
        pi(nil) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.