By Barnette D.W.

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The converse is equivalent to this: Given that XL + Z < y + Z, ZL + X < y + z, x + Z < YR + z, and x + Z < ZR + y, prove that XL < y and x < YR' B. Hmm. The converse would go through by induction - except that we might have a case with, say, XL + Z < y + Z but XL == y. Such cases would be ruled out by (T13), but ... A. But we need (T13) to prove (T14), and (T14) to prove (T13). And (T13) to prove (T12). B. We're going around in circles again. 67 A. Ah, but there's a way out, we'll prove them both together!

Allow me to give you a kiss for that. B. (smiling) The problem isn't completely solved, yet; we have to consider numbers like (0, {xj+d) and ({xi-d,0). But in the first case, we get the first-created number of Xl, X2, ... , Xj. And in the second case it's the first-created number of Xi, Xi+l, ... , Xm · A. What if the first-created number wasn't unique? I mean, what if more than one of the Xi, ... , Xj were created on that earliest day? B. Whoops ... No, it's okay, that can't happen, because the proof is still valid and it would show that the two numbers are both like each other, which is impossible.

Such cases would be ruled out by (T13), but ... A. But we need (T13) to prove (T14), and (T14) to prove (T13). And (T13) to prove (T12). B. We're going around in circles again. 67 A. Ah, but there's a way out, we'll prove them both together! We can prove the combined statement "(T13) and (T14)" by induction on the day-sum of (x, y, z)! B. (glowing) Alice, you're a genius! An absolutely gorgeous, tantalizing genius! A. Not so fast, we've still got work to do. We had better show that x -x =o. (T15) B.

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A 2-manifold of genus 8 without the W v-property by Barnette D.W.


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