Grobner basis: Basis for K-vector space. This is a problem from my Groner basis class. I am unsure about how to do this problem:

Let $A$ be an ideal of $R=K\left[x_{1},...,x_{n}\right]$, and let $G=\{g_{1},...,g_{t}\}$ be a Grobner basis for $A$. We want to prove that a basis for the $K$-vector space $R/A$ is
  $$\left\{X+A \ | \ X\in \mathbb{T}^{n}, lp(g_i)\; \text{does not divide}\; X, \forall i=1, \dots ,t\right\}.$$

lp: leading product, $\mathbb{T}^n$: the set of all power products, $f\xrightarrow{G}_{+}r$ : $f=\sum_{i}u_ig_i+r, u_i\in R, g_i\in G$ (reduction modulo $G$)
My attempt:
Suppose $A \leq R$ and $G=\{g_{1},...,g_{t}\}$ is a Grobner basis for $A$. If $f\in R$ then $f\xrightarrow{G}_{+}r$, $r$ reduced w.r.t $G$. Then, the coset representative for $R/A$ is 
$\left\{r+A \ | \ f \in R, f\xrightarrow{G}_{+}r\right\}$
Notice that $r=\sum_{i}c_iX_i$, where $c_i \in K, X_i\in \mathbb{T}^n$. We have that since $r$ reduced w.r.t G, $lp(g_i)$ not divide $lp(r), \forall i=1,...,t$ or $r=0$, then it means that $X=\max{X_i}=lp(r)$ does not divisible by $lp(g_i)$. Hence, a basis for the $K$-vector space $R/A$ is $$\left\{X+A \ | \ X\in \mathbb{T}^n, lp(g_i)\; \text{does not divide}\; X, \forall i=1,\dots...,t\right\}$$
I don't know whether my proof is correct or not. Really unsure about this.
 A: Let
$$B=\{X\in\mathbb{T}^n\mid X\text{ is not divisible by any }lp(g)\;,\;g\in G\}$$
$$B'=\{X+A\mid X\in B\}$$
As you noticed yourself every element $f+A$ can be represented as $r+A$ for some $r=\sum_ic_iX_i$ where the leading term of $r$ is not divisible by any leading terms of the members of $G$. When the leading term is not divisible by some monomials, you can conclude that the rest of terms are also not divisible by those monomials. Therefore not only $X$, but also all of $X_i$'s are in $B$. Now
$$f+A=r+A=(\sum_ic_iX_i)+A=\sum_ic_i(X_i+A)$$
This is the part you were attempting to do. What is left is to prove the linearity independent. Pick up a linear combination of elements of $B'$ and assume it is equal to zero.
$$\sum_ic_i(X_i+A)=0$$
Then $(\sum_ic_iX_i)+A=0$ which means $\sum_ic_iX_i\in A$. But then by definition of Grobner basis, the leading term of $\sum_ic_iX_i$ should be divisible by leading term of an element in $G$. It is impossible since the leading term of $\sum_ic_iX_i$ is one of the summands that has its monomial in $B$. Therefore the only possible case is that $c_i$'s all be zero. 
