The reason why he said that A->B is true is this : Because in the video we showed that ( A and ¬B ) is false And knowing that (A -> B) is equivalent to (¬A or B) , and, the negation of (¬A or B ) is (A and ¬B) which is false since the negation of (A->B) is false then (A->B) is true.
Unfortunately I'm using a Japanese book, which might probably not help you too much. If you're searching for the long term, I will ask my teacher. In case you're good with my videos, let me know about the topics you're interested in. In case I know something about it I will make a video
@@munteanionut3993 The video is correct. I forgot to mention a keyword here for people to search. I expected people to heard of this logic prior to watching this video. Meybe it helps if you look at 2:33 at the formula "A→B" and try to prove it yourself, given A and (¬)B as stated in the video.
@@munteanionut3993 Maybe also checking out how to use the "Horn scheme" might help. Or taking a look at my Prolog playlist (maybe overkill): ua-cam.com/video/6tfxnH_zFts/v-deo.html I couldn't find a video that explains just what happens there, sorry :/
The reason why he said that A->B is true is this :
Because in the video we showed that ( A and ¬B ) is false
And knowing that (A -> B) is equivalent to (¬A or B) , and, the negation of (¬A or B ) is (A and ¬B) which is false
since the negation of (A->B) is false then (A->B) is true.
Thank you!
Could you recommend any books for a begginer in this ?
Unfortunately I'm using a Japanese book, which might probably not help you too much. If you're searching for the long term, I will ask my teacher. In case you're good with my videos, let me know about the topics you're interested in. In case I know something about it I will make a video
@@lerneninverschiedenenforme7513 I would also be interested in an English resource is you ever found out :) Great series and thanks for making em!
So you are allowed to combine everything with everything?
No: 1:04 -> Take the result and combine it with anything
Does A -> B mean that A or ¬B has to be false?
See the other videos on this subject ua-cam.com/play/PLHbIZiLIDu-qbXqhvgkIjyIK0SbqYNcU8.html
Isn't: "(non-a OR b) AND a" actually equivalent to: "(non-a AND a) OR (b AND a)" which in turn simplifies to "b AND a"?
time?
@@lerneninverschiedenenforme7513 0:15... i think you shortened "(non-a OR b) AND a" wrong, but i am not sure, though. I am just asking.
@@munteanionut3993 The video is correct. I forgot to mention a keyword here for people to search. I expected people to heard of this logic prior to watching this video.
Meybe it helps if you look at 2:33 at the formula "A→B" and try to prove it yourself, given A and (¬)B as stated in the video.
@@munteanionut3993 Maybe also checking out how to use the "Horn scheme" might help. Or taking a look at my Prolog playlist (maybe overkill): ua-cam.com/video/6tfxnH_zFts/v-deo.html
I couldn't find a video that explains just what happens there, sorry :/
@@lerneninverschiedenenforme7513 ok, thank you!
THANK YOU!!!!!
I never expected any second person to need this stuff :D Glad I could help :)