abstract algebra – Doubt in computation of groebner basis and proving a set is a groebner basis.

I have a set $G$ , and I am trying to prove that the set is a groebner basis .

From chapter $2$ of Ideals,Varieties and Algorithm I have the following theorem,

A basis $G = {g_1,cdots, g_n }$ for an ideal I is a Gröbner basis if and only if $S(g_i, g_j) to_{G} 0 $ for all $i ne j$.

My ideal $I$ is generated by $y_1+ y_2*x_3 + y_3*x_3^2$,$y_1+ y_2*x_2 + y_3*x_2^2$,$y_1 + y_2*x_1+y_2*x_1^2$ and the monomial order is $(y_1>y_2>y_3>x_1>x_2>x_3)$ lexicographic ordering

The set $G$ is ${y_3*x_1^2*x_2 – y_3*x_1^2*x_3 – y_3*x_1*x_2^2 + y_3*x_1*x_3^2 + y_3*x_2^2*x_3 – y_3*x_2*x_3^2 , y_2*x_2 – y_2*x_3 + y_3*x_2^2 – y_3*x_3^2 , y_2*x_1 – y_2*x_3 + y_3*x_1^2 – y_3*x_3^2 , y_1+ y_2*x_3 + y_3*x_3^2}$

I am trying to claim is that this set $G$ is the groebner basis of $I$

I calculated the $6$, S-polynomials of the polynomials in $G$ and I have shown that the remainder obtained after dividing these (by the polynomials present in G) is $0$ .Is this process enough to conclude that $G$ is a groebner basis of $I$?Or should I be calculating the leading term of $S(g_i,g_j)$ where $i ne j$ and $g_i,g_j in G$ and then checking whether the leading term of this $S$-polynomial is present in the ideal generated by $langle LT(g_1),Lt(g_2) ,cdots Lt(g_n) rangle$?