A degree reverse lexicographic order $\prec$ is defined as follows:
Given the polynomial ring $R=K[x_1,...,x_n]$. Two monomials in $R$ have the order $x^u\prec x^v$, if $\deg(x^u)<\deg(x^v)$, or if $\deg(x^u)=\deg(x^v)$, then the last non-zero entry in $v-u$ is negative.
The question I'm asked is that given the polynomial ring $R=K[x_1,...,x_n]$ to show that if the degree condition is not present in the definition of a degree reverse lexicographic order, then the order is not a term order, i.e. $x^u\prec x^v\nrightarrow x^{u+r}\prec x^{v+r}$. Note that $u,v,r\in \mathbb{N}^n$.
Ignoring the degree condition, $x^u\prec x^v$ when the last non-zero entry of $v-u$ is negative. So if $u=(u_1,...,u_n)$ and $v=(v_1,...,v_n)$, then either $v_n-u_n<0$ or $v_n=u_n$, but I'm not sure where to go from here. I was thinking that I could assume that $\deg(x^v)<\deg(x^u)$ and then find $r=(r_1,...,r_n)$, and show that $x^{u+r}\not\prec x^{v+r}$, but I an't come up with an example.
Without the grading condition, $x^u \prec x^v$ if and only if $x^{u+r} \prec x^{v+r}$ still holds, since the last nonzero entry of $v-u$ is negative if and only if the last nonzero entry of $(v+r)-(u+r) = v-u$ is negative.
I expect that the thing that makes it not a term order is that you have $x^u \prec 1$ for any $u$, and in fact have infinite descending chains of the form $\ldots \prec x_i^3 \prec x_i^2 \prec x_i \prec 1$. Thus, this order isn't a well-ordering, which tends to be trouble if you want to do something productive with it.