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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        weight#_A(x1) = x1 + 28
        cons_A(x1,x2) = max{x1 - 4, x2 + 11}
        sum_A(x1,x2) = max{22, x1, x2 + 1}
        |0|_A = 6
        nil_A = 0
        s_A(x1) = max{13, x1}
  
    precedence:
    
      weight# = cons = sum = |0| = nil = s
    
    partial status:
  
      pi(weight#) = [1]
      pi(cons) = [2]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1,x2) = max{24, x1 + 6, x2 + 10}
        cons_A(x1,x2) = max{x1 + 1, x2 + 14}
        s_A(x1) = max{8, x1}
        |0|_A = 0
  
    precedence:
    
      sum# = cons = s = |0|
    
    partial status:
  
      pi(sum#) = [1, 2]
      pi(cons) = []
      pi(s) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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))

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(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1,x2) = max{0, x1 - 6}
        cons_A(x1,x2) = max{8, x1 + 4}
        s_A(x1) = max{9, x1 + 5}
  
    precedence:
    
      s > sum# = cons
    
    partial status:
  
      pi(sum#) = []
      pi(cons) = [1]
      pi(s) = []

The next rules are strictly ordered:

  p1

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