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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          weight#_A(x1) = ((0,1),(0,0)) x1 + (1,21)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,9)
          sum_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (13,3)
          |0|_A() = (1,2)
          nil_A() = (1,1)
          s_A(x1) = ((0,0),(1,1)) x1 + (0,1)
  
    precedence:
    
      weight# = sum > |0| > cons = s > nil
    
    partial status:
  
      pi(weight#) = []
      pi(cons) = [1]
      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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          sum#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (3,2)
          cons_A(x1,x2) = ((1,1),(0,0)) x2 + (2,1)
          s_A(x1) = x1 + (1,3)
          |0|_A() = (1,1)
  
    precedence:
    
      sum# = cons = s = |0|
    
    partial status:
  
      pi(sum#) = []
      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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          sum#_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,0)
          cons_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (1,0)
          s_A(x1) = ((0,0),(0,1)) x1 + (2,2)
  
    precedence:
    
      cons > sum# = s
    
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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