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. lexicographic path order with precedence:
    
      precedence:
      
        s > nil > sum > |0| > cons > weight#
      
      argument filter:
    
        pi(weight#) = 1
        pi(cons) = [2]
        pi(sum) = 2
        pi(|0|) = []
        pi(nil) = []
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > nil > sum > |0| > cons > weight#
      
      argument filter:
    
        pi(weight#) = []
        pi(cons) = 2
        pi(sum) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(s) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > nil > sum > |0| > cons > weight#
      
      argument filter:
    
        pi(weight#) = []
        pi(cons) = 2
        pi(sum) = [2]
        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. lexicographic path order with precedence:
    
      precedence:
      
        sum# > |0| > cons > s
      
      argument filter:
    
        pi(sum#) = [1, 2]
        pi(cons) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        sum# > |0| > s > cons
      
      argument filter:
    
        pi(sum#) = 2
        pi(cons) = 1
        pi(s) = 1
        pi(|0|) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        sum# > |0| > cons > s
      
      argument filter:
    
        pi(sum#) = [2]
        pi(cons) = 1
        pi(s) = 1
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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