Why are relations of degree 3 or less enough in a presentation of the polynomial current Lie algebra g[t]?

Let $mathfrak{g}$ be a finite dimensional simple Lie algebra over $mathbb{C}$.
The polynomial current Lie algebra $mathfrak{g}[t] = mathfrak{g} otimes mathbb{C} [t]$
has the bracket
$$[xt^r, yt^s] = [x,y] t^{r+s}$$
for $x,y in mathfrak{g}$. It is graded with deg$(t) = 1$.

If we set $h=0$ in Drinfeld's first presentation of the Yangian (given in Theorem 12.1.1 of Chari and Pressley's Guide to Quantum Groups) then we get a presentation of $U(mathfrak{g}[t])$
where the generators are the elements $x in mathfrak{g}$ and $J(x) = xt$ of $mathfrak{g}[t]$ with degree $=0,1$, and the relations all have degree of both sides less than $3$.

---

Specifically we require that all the relation in $mathfrak{g}$ are satisfied for the elements with degree 0, and
(for all $x,y, x_i, y_i, z_i in mathfrak{g}$ and complex numbers $lambda, mu$):

$$lambda xt + mu yt = (lambda x + mu y)t$$
$$[x, yt] = [x,y]t,$$
$$sum_i [x_i, y_i] = 0 implies sum_i [x_i t, y_i t ] = 0$$
$$ sum_i [[x_i, y_i], z_i] = 0 implies sum_i [[x_i t, y_i t], z_i t]=0$$
Then assuming that all the relations of degree less than or equal to $3$ hold is enough to get the remaining ones.
The elements $xt^2, xt^3, ldots$ are defined inductively.
This can be proved by induction, using the Serre presentation of the finite-dimensional Lie algebra and then checking all the required relations in several cases.
But even in the $mathfrak{sl}_2$ case the argument is laborious.

Is there a better way of seeing that one needs only relations of degree less than three in order to get the rest?