Def. let be $\preceq_A$ a total order, $(a_1,a_2,...,a_n),(b_1,b_2,...,b_n) \in A^n$, $(a_1,a_2,...,a_n) \leq^d (b_1,b_2,...,b_n)$ if one and only one of the following holds is true: $$a_1 \prec_A b_1$$$$a_1=b_1 \wedge a_2 \prec_A b_2$$$$a_1=b_1 \wedge a_2=b_2 \wedge a_3 \prec_A b_3$$$$.$$$$.$$$$.$$$$a_1=b_1 \wedge a_2=b_2 \wedge ... \wedge a_{n-1}=b_{n-1} \wedge a_n \prec_A b_n$$with $a_i \prec_A b_i \equiv a_i \preceq_A b_i \wedge a_i \neq b_i$
Is it correct? Thanks in advance
To reflect that $\le$ is associated with $\preceq$ and $<$ with $\prec$ you could use $$\begin{align*}a\prec b & :\Leftrightarrow a\preceq b \wedge a\neq b\\ A<^d B & :\Leftrightarrow \exists k: a_i = b_i \forall i<k \quad \wedge \quad a_k \prec b_k\\ A\le^d B&:\Leftrightarrow A<^d B \vee A=B\end{align*}$$
If you want to only define $\le^d$ but equivalent to the above (i.e. eliminate usage of $\prec, <$), write this:
$$A\le^d B:\Leftrightarrow A = B \vee (\exists k: a_i=b_i \forall i <k\quad \wedge\quad a_k \ne b_k \wedge a_k \preceq b_k)$$ $k$ could be called the "common prefix length" or something.