Timeline for Using logical equivalence to prove PL statements
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| May 17, 2020 at 1:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 18, 2020 at 0:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 17, 2019 at 23:38 | comment | added | Ethan | @user64494 My free online cloud can't run it too $\dots$ but I tried use it to prove two Consensus theorems, where Identity, Domination, Negation laws play a important roles in the proof: \begin{align} &\text{1. or[or[and[x,y],and[not[x],z]],and[y,z]]==or[and[x,y],and[not[x],z]]]}\\ &\text{2. and[and[or[x,y],or[not[x],z]],or[y,z]]==and[or[x,y],or[not[x],z]]}\\ \end{align} and both axiom works, use True, Flase seems make no difference from T,F, but thanks for the suggestion. | |
| Dec 15, 2019 at 18:00 | history | tweeted | twitter.com/StackMma/status/1206272854247059456 | ||
| Dec 15, 2019 at 7:57 | comment | added | user64494 | I think True and False in your axioms make problems. Try to replace these by T and F. My comp is too weak to realize my suggestion (so does my Mathematica Online). | |
| Dec 15, 2019 at 5:07 | answer | added | ShyPerson | timeline score: 1 | |
| Nov 17, 2019 at 1:58 | history | edited | J. M.'s missing motivation | edited tags | |
| Nov 10, 2019 at 2:33 | history | edited | Ethan | CC BY-SA 4.0 | added 31 characters in body |
| Nov 10, 2019 at 2:11 | history | edited | Ethan | CC BY-SA 4.0 | added 31 characters in body |
| Nov 9, 2019 at 22:59 | history | edited | Ethan | CC BY-SA 4.0 | improved formatting |
| Nov 9, 2019 at 22:53 | history | edited | Ethan | CC BY-SA 4.0 | added 31 characters in body |
| Nov 9, 2019 at 18:59 | history | edited | Ethan | CC BY-SA 4.0 | added 31 characters in body |
| Nov 9, 2019 at 17:17 | comment | added | Ethan | @LukasLang Thanks! | |
| Nov 9, 2019 at 15:48 | comment | added | Lukas Lang | the third-to-last axiom seems wrong: $\lnot(p\lor q)\equiv\lnot p\lor\lnot q$ should be $\lnot(p\lor q)\equiv\lnot p\land\lnot q$ | |
| Nov 9, 2019 at 7:32 | comment | added | mikado | You could build a truth table (true and false) for each value of p, q and r. That might identify any invalid axioms. | |
| Nov 9, 2019 at 7:00 | review | First posts | |||
| Nov 9, 2019 at 14:03 | |||||
| Nov 9, 2019 at 6:59 | history | asked | Ethan | CC BY-SA 4.0 |