David Cox, Ideals, varieties and algorithms page 69 exercise 11.
Let $f=q_1f_1+\cdots+q_sf_s+r$ be an expression produced by the division algorithm. Let $\text{LM}(f_i)=x^{\alpha(i)}$ and we define $$ \Delta_1=\alpha(1)+\mathbb{N}^n,\Delta_2=(\alpha(1)+\mathbb{N}^n)-\Delta_1,\ldots,\Delta_s=(\alpha(s)+\mathbb{N}^n)-(\displaystyle {\cup_{i=1}^s \Delta_i}),\overline{\Delta}=\mathbb{N}^n-\left(\displaystyle {\cup_{i=1}^s \Delta_i}\right).$$
Given: for every $i$, every monomial $x^\beta$ in $q_i$ satisfies $\beta+\alpha(i)\in \Delta_i$, and every monomial $x^\gamma$ in $r$ satisfies $\gamma\in \overline{\Delta}.$
Show that there is exactly one expression $f=q_1f_1+\cdots+q_sf_s+r$ satisfying the Given.
Solution:suppose that $f=q_1f_1+\cdots+q_sf_s+r=f=q_1'f_1+\cdots+q_s'f_s+r'$, then $r=f-q_1f_1-\cdots-q_sf_s=q_1'f_1+\cdots+q_s'f_s+r'-q_1f_1-\cdots-q_sf_s$. Since no term of $r$ is divisible by $\text{ LT}(f_i)$, then $(q_1'-q_1)f_1+\cdots+(q_s'-q_s)f_s$ must vanish. Hence $r=r'$.
But I do not have any idea to show that $q_i=q_i'$.
