A Complete characterization of local monomials

[Gaussian97]

TL;DR Summary

If $[A(x), j_{\{\lambda\}}(y)]=0$ for $(x-y)^2<0$ then $A(x)$ is a local polynomial.

Hi, I'm reading Zavialov's book on QFT and there's a statement there I was interested in finding how to prove it. The statement is as follows:

If $[A(x), j_{\{\lambda\}}(y)]=0$ for $(x-y)^2<0$ then $A(x)$ is a local polynomial.

The relevant definitions are:
$$A = \sum_n \int A_n(x_1,\ldots, x_n) :\phi(x_1)\cdots\phi(x_n): dx_1 \cdots dx_n$$
$$j_{\{\lambda\}}(x)=:\phi_{(\lambda_1)}(x)\cdots \phi_{(\lambda_n)}(x):$$
$$\phi_{(\lambda_1)}(x) = \left(\frac{\partial}{\partial x}\right)^{(\lambda_i)}\phi(x)$$
and a local polynomial is just a linear combination of $j_{\{\lambda\}}(x)$.

Does anyone have a proof of this fact?

This are the main thoughts I came up trying to prove the statement:

The first thing is that the book does not explicitly define $A(x)$, but I think it is probably defined by the natural
$$A(x) = \sum_n \int A_n(x,x_1,\ldots, x_n) :\phi(x_1)\cdots\phi(x_n): dx_1 \cdots dx_n$$
From here the first thing I thought was that the commutator $[A(x), j(y)]$ reduces to know the value of the commutators
$$[:\phi(x_1)\cdots \phi(x_n):, j(y)]$$
And since the derivatives can be extracted from the commutator at the end we need to compute
$$[:\phi(x_1)\cdots \phi(x_n):, :\phi(y_1)\cdots \phi(y_m):]$$
Now, using the Wick theorem we can get rid of the normal ordering, and using the relation $[A,BC]=B[A,C]+[A,B]C$, we should really be able to reduce everything to the commutator $[A(x), \phi(y)]$. So, either I'm neglecting something important or the assumption $[A,j]=0 \forall j$ seems to be innecesary, just needing $[A,\phi]=0$, no?

Then we can write the commutators
$$[\phi(x_1)\cdots \phi(x_n), \phi(y)] = \sum_{i=1}^n [\phi(x_i), \phi(y)] \phi(x_1)\cdots\phi(x_{i-1})\phi(x_{i+1})\cdots\phi(x_n)$$

My idea then is trying to prove that this sum cannot vanish unless each particular term vanishes. We know that $[\phi(x_i), \phi(y)]$ vanishes when $(x_i-y)^2<0$ so the condition $[A,\phi]=0$ would imply that the functions $A_n(x,x_1,\ldots, x_n)$ must vanish whenever $(x_i-y)^2\geq 0$.
Since this must be true for all $y$ such that $(x-y)^2<0$ this implies that $A_n(x,x_1,\ldots, x_n)$ must be zero whenever $x_i\neq x$ and therefore all the functions $A_n(x,x_1,\ldots, x_n)$ would be linear combinations of $\delta(x-x_i)$ and their derivatives. Which would prove that $A(x)$ is a local polynomial.

I don't know if anyone knows if this is the right approach. The part that I think I'm more distant to actually prove (with some rigour) is to prove that indeed the sum cannot be zero unless each individual term, which seems plausible to me. But I have no clue.

If anyone knows the proof or wants to share any thoughts about it, I'd like to hear you.