The Weyl algebra is $ A=k\langle x,y\rangle/\langle yx-xy-1\rangle $ where $k$ is a field, $k\langle x,y\rangle$ is the free algebra, and $yx-xy-1$ is the defining relation. A basis for $A$ is $\{ x^i y^j|i,j\geq0\}$. What I am concerned about here is how to prove that elements of this set are linearly independent. Can we reason by contradiction that if $\sum c_{ij}x^i y^j=0$, then (after rewriting $\sum c_{ij}x^i y^j=\sum f_j(x)y^j$), $f_j(x)=0,\forall j$, thus $c_{ij}=0,\forall i,j$. This is a bit weird, I feel.
Another related question is about the $q$-Weyl algebra $ B=k\langle x,y\rangle/\langle xx^{-1}-1,x^{-1}x-1,yy^{-1}-1,y^{-1}y-1,yx-qxy \rangle $ where $q\in k$. How to show that that a basis for $B$ is $\{x^i y^j|i,j\in \mathbb{Z}\}$? Can someone give me some hints on these two problems? Thank you very much!