Timeline for Arriving at the same result with the opposite hypotheses
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2019 at 21:43 | comment | added | Jirka Hanika | @ToothpickAnemone - The answer was following (a special case of) the pattern posed in the question, not vice versa. Joel has explored the special case of P = Q. | |
| Jun 4, 2019 at 20:15 | comment | added | Toothpick Anemone |
I do not think that the proof structure being discussed by the question asker follows your pattern. Their pattern is [(P) --> (Q)] and [(not P) --> (Q)] therefore, (Q). That is standard proof by cases. You did something different.
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| Jun 3, 2019 at 15:35 | comment | added | user141414 | I appreciate your shift of perspective. It is actually pretty interesting. Still, you addressed a very different question. If I could downvote, I would (but I also thank you for this insight, even if it was not presented in the appropriate place!). | |
| Jun 3, 2019 at 15:28 | comment | added | Joel David Hamkins | I took the central phenomenon of your question to be the situation where one statement follows from another and also from its negation. While that might seem unusual or even amazing, as you say, the point of my answer is that actually this happens quite frequently in mathematics. | |
| Jun 3, 2019 at 15:25 | comment | added | user141414 | I do not see how this addresses the question. The question specifically asked about "major, notorious conjectures in mathematics". A somewhat subjective definition but you probably agree that not every statement $S$ belongs to that category. Actually, now that I think of it $S$ would simultaneously have to be proven and to be a conjecture so I think this answer deserves some further clarification. | |
| S Jun 3, 2019 at 15:24 | history | answered | Joel David Hamkins | CC BY-SA 4.0 | |
| S Jun 3, 2019 at 15:24 | history | made wiki | Post Made Community Wiki by Joel David Hamkins |