YES

TRS:

  active(first(0(),X)) -> mark(nil())
  active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
  active(from(X)) -> mark(cons(X,from(s(X))))
  mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
  mark(0()) -> active(0())
  mark(nil()) -> active(nil())
  mark(s(X)) -> active(s(mark(X)))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(from(X)) -> active(from(mark(X)))
  first(mark(X1),X2) -> first(X1,X2)
  first(X1,mark(X2)) -> first(X1,X2)
  first(active(X1),X2) -> first(X1,X2)
  first(X1,active(X2)) -> first(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  cons(mark(X1),X2) -> cons(X1,X2)
  cons(X1,mark(X2)) -> cons(X1,X2)
  cons(active(X1),X2) -> cons(X1,X2)
  cons(X1,active(X2)) -> cons(X1,X2)
  from(mark(X)) -> from(X)
  from(active(X)) -> from(X)

linear polynomial interpretations on N:

  active_A(x1) = x1
  active#_A(x1) = x1 + 2
  first_A(x1,x2) = x1 + x2 + 4
  first#_A(x1,x2) = 7
  0_A = 3
  0#_A = 0
  mark_A(x1) = x1
  mark#_A(x1) = x1 + 4
  nil_A = 4
  nil#_A = 1
  s_A(x1) = x1 + 1
  s#_A(x1) = 0
  cons_A(x1,x2) = x1 + 2
  cons#_A(x1,x2) = 5
  from_A(x1) = x1 + 5
  from#_A(x1) = 1

precedence:

  active > mark > first = from > nil = s = cons > 0