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Is the following statement true? If so, can someone advise on how to prove it?

Let A = $(S_1,\prec_1)$ be a partially-ordered set and let B = $(S_1,\prec_2)$ be a totally-ordered set

Let $\preccurlyeq$ be the lexicographic order on $A \times B$.

Then $\preccurlyeq$ is an total ordering on $A \times B$.

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It’s false in general. Let $S_1=\{a,b\}$, where $a$ and $b$ are incomparable with respect to $\prec_1$, and $a\prec_2b$. Then $\langle S_1\times S_1,\preccurlyeq\rangle$ is not a total order, as can be seen from its Hasse diagram:

             <a,b>        <b,b>  
               |            |  
             <a,a>        <b,a>
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  • $\begingroup$ I am sorry but I made several typos; I will correct the question. $\endgroup$ – S0rin Apr 25 '15 at 19:22
  • $\begingroup$ @S0rin: I’ve adjusted my answer to match the corrected question. $\endgroup$ – Brian M. Scott Apr 25 '15 at 19:33

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