As I don't know what theorem 9 tries to say, I can't follow the highlighted statement in the proof of theorem 14. Could you please help explain theorem 9 in plain English to a layman like me, and why it leads to that statement d + e <= e?
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A partial ordering is a binary relation which reflexive, anti-symmetric and transistive. What matters here is that it is anti-symmetric: if $e\leqq d+e$ and if $d+e\leqq d$, then $e=d+e$.
answered Nov 26, 2019 at 10:47
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$\begingroup$ Thanks for your explanation, Jose. I can see e <= d + e but how can d + e <= e? $\endgroup$– NemoCommented Nov 26, 2019 at 11:03
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1$\begingroup$ Because $d\leqq e\implies d+e\leqq e+e=e$. $\endgroup$ Commented Nov 26, 2019 at 11:05
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$\begingroup$ Oh, I see. The brief highlighted sentence of the proof threw me off as I thought it was a statement/given fact. Thank you! One more question: why is it called partial? What does partial refer to? $\endgroup$– NemoCommented Nov 26, 2019 at 11:12
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1$\begingroup$ A "total" ordering is one in which any two different elements are comparable, and one of the two is better. A "partial" ordering is not so fussy. Among two elements, at most one is better, but it might also happen that the two elements can't be compared and we can't say which is better. $\endgroup$– MJDCommented Nov 26, 2019 at 12:19
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$\begingroup$ Thanks, @MJD. What I find difficult to understand set theory is that it often gives no example to abstract concepts, thus I can't visualise and remember their definitions. But theorem 9 talks about the relation <=, not elements, so how can a relation be partially ordering? Or does it mean we can't compare between a <= b and d <= e? $\endgroup$– NemoCommented Nov 27, 2019 at 0:12