Timeline for Is the disjunction of these two false statements true?
Current License: CC BY-SA 4.0
23 events
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| May 17, 2018 at 18:04 | vote | accept | Randy Randerson | ||
| May 17, 2018 at 14:40 | answer | added | Matt | timeline score: 8 | |
| S May 17, 2018 at 9:14 | history | suggested | CommunityBot | CC BY-SA 4.0 |
That's not how Stack Exchange works
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| May 17, 2018 at 8:44 | review | Suggested edits | |||
| S May 17, 2018 at 9:14 | |||||
| May 17, 2018 at 8:08 | answer | added | kutschkem | timeline score: -1 | |
| May 17, 2018 at 6:34 | history | edited | Randy Randerson | CC BY-SA 4.0 |
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| May 17, 2018 at 6:21 | comment | added | Mauro ALLEGRANZA | Regarding the Edit, you are again wrong: they are NOT both false. If you flip the coin and shows head, then the first one is $T \to F$, which is FALSE, but the second is $F \to T$, which is TRUE. | |
| May 17, 2018 at 6:18 | comment | added | Mauro ALLEGRANZA | But obviously $\forall x (x \ge 3) \lor \forall (x < 3)$ (which is FALSE) is NOT equiv to : $\forall x (x \ge 3 \lor x <3)$ (which is TRUE). And this is a perfect counterexample showing that the universal quantifier does not distribute over disjunction. | |
| May 17, 2018 at 6:16 | comment | added | Mauro ALLEGRANZA | The first one is equiv to $\forall x ((x \ge 3) \lor (x \ge 3))$ which trivially is equiv to $\forall x (x \ge 3)$. The second one is equiv to $\forall x ((x < 3) \lor (x < 3))$ which again amounts to: $\forall x (x < 3)$. | |
| May 17, 2018 at 6:12 | review | Close votes | |||
| May 22, 2018 at 3:03 | |||||
| May 17, 2018 at 3:25 | comment | added | jdods | @RandyRanderson, also, This question is perfectly suited for mathSE, in my opinion. | |
| May 17, 2018 at 3:16 | answer | added | Graham Kemp | timeline score: 12 | |
| May 17, 2018 at 3:14 | comment | added | jdods | With your coin example, imagine flipping the coin and it comes up heads. Then "if it is tails, then it is heads" is true vacuously. "False implies true" is true. If the coin hasn't been flipped yet then no truth values are assigned. I think it is confusing since it is not intuitive, but thinking of "if it is tails, then it is heads" as intuitively false is built on the idea of the premise "it is tails" being true. And that is correct! A coin that shows tails certainly does not show heads! | |
| May 17, 2018 at 3:11 | comment | added | Dan Brumleve | Consider "if the moon is made of cheese, then the coin shows tails." It's true, since the moon isn't made of cheese. So "if the coin shows heads, then the coin shows tails" is also true if the coin doesn't show heads. | |
| May 17, 2018 at 3:10 | comment | added | Randy Randerson | That makes sense | |
| May 17, 2018 at 3:07 | comment | added | jdods | I think the concept of vacuous truth is relevant here. Either $A$ or its negation is true thus it is always the case that one of the implication statements $(\sim A) \rightarrow A$ or $A \rightarrow (\sim A) $ is vacuously true. | |
| May 17, 2018 at 3:03 | history | edited | Randy Randerson | CC BY-SA 4.0 |
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| May 17, 2018 at 2:41 | history | edited | Randy Randerson | CC BY-SA 4.0 |
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| May 17, 2018 at 2:35 | answer | added | hmakholm left over Monica | timeline score: 27 | |
| May 17, 2018 at 2:34 | answer | added | Dan Brumleve | timeline score: 4 | |
| May 17, 2018 at 2:27 | history | edited | Randy Randerson | CC BY-SA 4.0 |
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| May 17, 2018 at 2:24 | answer | added | vadim123 | timeline score: 5 | |
| May 17, 2018 at 2:18 | history | asked | Randy Randerson | CC BY-SA 4.0 |