Yes, you could use the term lexicalThe lexical order here. Usually, if one hasof a well-ordered ordered list of orders $P_\alpha$, then one may define$\langle X_\alpha,\leq_\alpha\rangle$ is the lexical order on the product of the underlying space by$\Pi_\alpha X_\alpha$ placing $\vec x$$s$ before $\vec y$$t$ if $s_\alpha\lt_\alpha t_\alpha$ on the firstleast difference $x_\alpha$ is before $y_\alpha$ in $P_\alpha$coordinate. In
But this is not your casesituation, you don't havesince all your orders, but merely pre-orders live on the same set (since twothe set of sheep could have the same age, etc.).
But in your case, and you don't wantseek an order on that set, not on the product spaceset. That is, since you want to order the individual sheep, not sequences of sheep. All your orders have the same underlying set (the set of sheep), and so this isn't the usual meaning ofBut if you simply restrict the lexical order. One way to the constant sequences, you get what you want, and for this reason, I think about it is that you are restrictingfine to call it the lexical order to the constant sequences.
If $P_\alpha$Another difference is that you don't actually have orders, but pre-orders, since two sheep can have the same age, and so on. In this general setting, you have a set $X$ with a well-ordered seqeuncesequence of pre-orders on a set $X$$\leq_\alpha$, we mayand you define the order on $X$ by placing $x$ before $y$$a\leq b$ if the least index $\alpha$ where theyfor which $a$ and $b$ are not equivalent has $x$ before(if any such $y$ in$\alpha$ exists) has $P_\alpha$$a\lt_\alpha b$.
It is important that the setlist of orders is well-ordered, since otherwise there may be no such least difference coordinate $\alpha$, and the idea will run aground. But any finite list of criteria is of course well-ordered, as is any $\omega$-sequence of orders.