YES

We show the termination of the TRS R:

  sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
  sum(cons(|0|(),x),y) -> sum(x,y)
  sum(nil(),y) -> y
  weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
  weight(cons(n,nil())) -> n

-- SCC decomposition.

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

p1: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))
p2: sum#(cons(|0|(),x),y) -> sum#(x,y)
p3: weight#(cons(n,cons(m,x))) -> weight#(sum(cons(n,cons(m,x)),cons(|0|(),x)))
p4: weight#(cons(n,cons(m,x))) -> sum#(cons(n,cons(m,x)),cons(|0|(),x))

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
r5: weight(cons(n,nil())) -> n

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1, p2}


-- Reduction pair.

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

p1: weight#(cons(n,cons(m,x))) -> weight#(sum(cons(n,cons(m,x)),cons(|0|(),x)))

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
r5: weight(cons(n,nil())) -> n

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            weight#_A(x1) = x1 + (2,2)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (1,1)
            sum_A(x1,x2) = x2
            |0|_A() = (5,2)
            nil_A() = (1,1)
            s_A(x1) = ((1,0),(1,1)) x1
    
      precedence:
      
        weight# = cons = sum = |0| = nil = s
      
      partial status:
    
        pi(weight#) = []
        pi(cons) = []
        pi(sum) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            weight#_A(x1) = (2,1)
            cons_A(x1,x2) = (3,3)
            sum_A(x1,x2) = x2 + (4,4)
            |0|_A() = (1,2)
            nil_A() = (1,1)
            s_A(x1) = ((0,0),(1,0)) x1 + (1,6)
    
      precedence:
      
        weight# = cons = sum = |0| = s > nil
      
      partial status:
    
        pi(weight#) = []
        pi(cons) = []
        pi(sum) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))
p2: sum#(cons(|0|(),x),y) -> sum#(x,y)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
r5: weight(cons(n,nil())) -> n

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (2,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (4,14)
            s_A(x1) = ((1,0),(1,0)) x1 + (3,13)
            |0|_A() = (1,2)
    
      precedence:
      
        s = |0| > sum# = cons
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(s) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1,x2) = (0,0)
            cons_A(x1,x2) = (2,1)
            s_A(x1) = (1,0)
            |0|_A() = (1,1)
    
      precedence:
      
        |0| > cons = s > sum#
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = []
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: sum#(cons(|0|(),x),y) -> sum#(x,y)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
r5: weight(cons(n,nil())) -> n

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: sum#(cons(|0|(),x),y) -> sum#(x,y)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: weight(cons(n,cons(m,x))) -> weight(sum(cons(n,cons(m,x)),cons(|0|(),x)))
r5: weight(cons(n,nil())) -> n

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,2)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (1,0)
            |0|_A() = (1,1)
    
      precedence:
      
        cons > sum# = |0|
      
      partial status:
    
        pi(sum#) = [2]
        pi(cons) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1,x2) = x2
            cons_A(x1,x2) = (1,1)
            |0|_A() = (1,1)
    
      precedence:
      
        sum# = |0| > cons
      
      partial status:
    
        pi(sum#) = [2]
        pi(cons) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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