YES

We show the termination of R/S, where R is

  app(nil,k) -> k
  app(l,nil) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil)) -> cons(x,nil)
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
  plus(0,y) -> y
  plus(s(x),y) -> s(plus(x,y))
  sum(plus(cons(0,x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
  pred(cons(s(x),nil)) -> cons(x,nil)

and S is:

  cons(x,cons(y,l)) -> cons(y,cons(x,l))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  app#(cons(x,l),k) -> app#(l,k)
  sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
  sum#(cons(x,cons(y,l))) -> plus#(x,y)
  sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
  sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
  sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
  plus#(s(x),y) -> plus#(x,y)
  sum#(plus(cons(0,x),cons(y,l))) -> pred#(sum(cons(s(x),cons(y,l))))
  sum#(plus(cons(0,x),cons(y,l))) -> sum#(cons(s(x),cons(y,l)))

and Q consists of the rules:

  app(nil,k) -> k
  app(l,nil) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil)) -> cons(x,nil)
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
  plus(0,y) -> y
  plus(s(x),y) -> s(plus(x,y))
  sum(plus(cons(0,x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
  pred(cons(s(x),nil)) -> cons(x,nil)
  cons(x,cons(y,l)) -> cons(y,cons(x,l))

The weakly monotone algebra (N^3, >_lex) with

  app#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2 + 2, z1)
  cons_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 2, y1 + y2 + 2, z2 + 2)
  sum#_A((x1,y1,z1)) = (x1, y1, z1)
  plus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, 3, 5)
  plus#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2 + 5, z1 + z2)
  app_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, y1 + y2 + 1, z1 + z2 + 1)
  sum_A((x1,y1,z1)) = (x1, y1, 3)
  s_A((x1,y1,z1)) = (x1 + 1, y1, z1)
  0_A = (1, 1, 1)
  pred#_A((x1,y1,z1)) = (x1, 0, 6)
  nil_A = (1, 1, 1)
  pred_A((x1,y1,z1)) = (x1, y1 + 1, z1 + 1)

strictly orients the following dependency pairs:

  app#(cons(x,l),k) -> app#(l,k)
  sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
  sum#(cons(x,cons(y,l))) -> plus#(x,y)
  sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
  sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
  sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
  plus#(s(x),y) -> plus#(x,y)
  sum#(plus(cons(0,x),cons(y,l))) -> pred#(sum(cons(s(x),cons(y,l))))
  sum#(plus(cons(0,x),cons(y,l))) -> sum#(cons(s(x),cons(y,l)))

We remove them from the set of dependency pairs.

No dependency pair remains.