Let $(M,\leq)$ be a totally ordered set, a word $w$ on $M$ is a finite series of elements from W, such that $$ w = w_{1}.....w_{n}, \hspace{1cm} w_1,....,w_n \in M, n \in \mathbb{N} $$ The positive integer n is the length of the word $w$. We define a relation $\preceq$ on the set of all words with the length $n$, for two words $v=v_1....v_n$ and $w=w_1...w_n$ the relation $v \prec w$ holds iff $\exists k \leq n$ such that $v_j = w_j$ for $1 \leq j < k$ and $v_k < w_k$. Show that $\preceq$ is a order relation. Is it also a total order?
I thought of an example myself, e.g. the words v="advice" and w="advise", for both n=6 and $v_5$ < $w_5$
So I need to show 1))reflexivity 2)transitivity 3)antisymmetric 4) total order axiom
1)reflexivity:
$v\preceq v$ holds, since $\forall j$ with $1 \leq j \leq n$ $v_j=v_j$ (can I say it like this or do I need a formal proof?)
2) antisymmetry
suppose that $v \prec w$ $\And$ $w \prec v$, then $v_k < w_k$ $\land$ $w_k < v_k$. I am stuck how to disprove this.
3) transitivity for $v \prec w$ $\land$ $w \prec x$ $\rightarrow$ $v \prec x$ hence $\exists k \leq n$: (or could the k also be different?) $v_{k} < w_{k}$ $\land$ $w_{k} < x_{k}$. I have problems formally proving this and I also need to prove/disprove the total order rule: $(v \prec w)$ $\lor$ $(w \prec v)$
1) The problem statement explicitly defines onl $\prec$. I assume that $x\preceq y$ is implicitly defined as $x\prec y\lor x=y$. Then by the "${}\lor x=y$" part, this is automatically reflexive, no matter how weird $\prec$ is defined.
2) Suppose $v\prec w$, i.e., for some $1\le k\le n$, we have $v_k<w_k$ and $v_i=w_i$ for $1\le i<k$. If also $w\prec v$, then for some $1\le k'\le n$, we have $w_{k'}<v_{k'}$ and $v_i=w_i$ for $1\le i<k'$. If $k'<k$, then we have $v_{k'}=w_{k'}$ from the first, but $w_{k'}<v_{k'}$ from the second, which is absurd. Similarly, $k<k'$ leads to a contradiction. We conclude that $k=k'$. But then the first gives $v_k<w_k$ and the second $w_k<v_k$, contradiction. In summary, $v\prec w\land w\prec v$ cannot occur.
3) Yes, the $k$'s may differ (cf. the full proof for 2 above). From the given there exists $1\le k\le n$ with $v_k<w_k$ and $v_i=w_i$ for $1\le i<k$. And there exists $1\le k'\le n$ with $w_{k'}<x_{k'}$ and $w_i=x_i$ for $1\le i<k'$. If $k\le k'$, we conclude that $v_k<w_k\le x_k$ and $v_i=w_i=x_i$ for $1\le i<k$ and so $v\prec x$. Simliarly if $k>k'$, we conclude that $v_{k'}=w_{k'}< x_{k'}$ and $v_i=w_i=x_i$ for $1\le i<k'$ and so again $v\prec x$.
Assume $v\ne w$, i.e., there exists $1\le j\le n$ with $v_j\ne w_j$, i.e., $A:=\{\,j\in\Bbb N\mid 1\le j\le n, v_j\ne w_j\,\}$ is a non-empty subset of $\Bbb N$. Accordingly, we can let $k=\min A$. Then $1\le k\le n$ and by minimality, $v_i=w_i$ for $1\le i<k$, and finally as $v_k<w_k$ or $w_k<v_k$, we see that $v\prec w$ or $w\prec v$.