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:
    
        max/plus interpretations on natural numbers:
        
          weight#_A(x1) = max{9, x1}
          cons_A(x1,x2) = max{12, x1 + 10, x2 + 4}
          sum_A(x1,x2) = x2 + 4
          |0|_A = 2
          nil_A = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        |0| > cons > weight# = sum > nil = s
      
      partial status:
    
        pi(weight#) = [1]
        pi(cons) = []
        pi(sum) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          weight#_A(x1) = x1 + 8
          cons_A(x1,x2) = 9
          sum_A(x1,x2) = 1
          |0|_A = 7
          nil_A = 0
          s_A(x1) = max{10, x1}
    
      precedence:
      
        cons = |0| = nil = s > sum > weight#
      
      partial status:
    
        pi(weight#) = [1]
        pi(cons) = []
        pi(sum) = []
        pi(|0|) = []
        pi(nil) = []
        pi(s) = [1]
    

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:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1,x2) = max{8, x1 + 4, x2 + 2}
          cons_A(x1,x2) = max{x1 + 2, x2 + 4}
          s_A(x1) = x1
          |0|_A = 2
    
      precedence:
      
        sum# = cons = s = |0|
      
      partial status:
    
        pi(sum#) = [1, 2]
        pi(cons) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1,x2) = x1 + 6
          cons_A(x1,x2) = max{x1 + 2, x2}
          s_A(x1) = x1 + 3
          |0|_A = 0
    
      precedence:
      
        sum# = cons = s = |0|
      
      partial status:
    
        pi(sum#) = [1]
        pi(cons) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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