11:35 I thinks the operators are wrong. The negation of p ∧ q is ¬(p ∧ q) and not ¬(p ∨ q) as stated in the video. I think the correct negation should be ¬(p ∧ q) ≡ (¬p ∨ ¬q)
Please see the updated videos at 1.3.1: ua-cam.com/video/tj_98IO-lCk/v-deo.html ("Proving" Logical Equivalances) 1.3.2: ua-cam.com/video/aXobNQArW64/v-deo.html (Key Logical Equivalences Including DeMorgan's Law) 1.3.3: ua-cam.com/video/u5y1zte2w_4/v-deo.html (Constructing New Logical Equivalences) The full playlist for Discrete Math I (Rosen, Discrete Mathematics and Its Applications, 7e) can be found at ua-cam.com/play/PLl-gb0E4MII28GykmtuBXNUNoej-vY5Rz.html
I did the same and it is satisfiable. I am in this video because my professor posts videos from this account to study. Obviously the pandemic has made his job easier and our job harder and he gets paid from us paying the tuition.
Years later and your vides are still valid, honestly you are a life saver
11:35 I thinks the operators are wrong. The negation of p ∧ q is ¬(p ∧ q) and not ¬(p ∨ q) as stated in the video. I think the correct negation should be ¬(p ∧ q) ≡ (¬p ∨ ¬q)
what happened at 23:03 ?
Haha, not sure. I'll have to rerecord that slide. Thanks for letting me know!
I laughed so hard multiple times lol
Please leave that part in. It made my night
@@SawFinMath You're so real for that 😂
Thank you so much for the easy explanation of lectures. It was so helpful for me and I appreciate your work.
Good Luck
11:35 Should be !(P && Q) -> (!P OR !Q)
15:06
You just dropped the addition sign , why?
Please see the updated videos at
1.3.1: ua-cam.com/video/tj_98IO-lCk/v-deo.html ("Proving" Logical Equivalances)
1.3.2: ua-cam.com/video/aXobNQArW64/v-deo.html (Key Logical Equivalences Including DeMorgan's Law)
1.3.3: ua-cam.com/video/u5y1zte2w_4/v-deo.html (Constructing New Logical Equivalences)
The full playlist for Discrete Math I (Rosen, Discrete Mathematics and Its Applications, 7e) can be found at ua-cam.com/play/PLl-gb0E4MII28GykmtuBXNUNoej-vY5Rz.html
at 27:30 can't we assign p = F, ~p = T, q = F, ~q = T, r = T, ~r = F which can make the whole expression satisfiable?
Thank you so much for your nice and easy explanations
What's the name of the rule used at 20:59?
20:55 - I think it is called Material Implication: en.wikipedia.org/wiki/Material_implication_(rule_of_inference)
At 27:20 , for problem #3, wouldn't the solution be satisfied if p = T; q = T; r = F?
Never mind.
I think for no 3 if you make P = T, Q = F, R = F. it would be satisfiable
I did the same and it is satisfiable. I am in this video because my professor posts videos from this account to study. Obviously the pandemic has made his job easier and our job harder and he gets paid from us paying the tuition.
23:03 I love you😂❤️❤️❤️❤️
prof flexing she got a pool eh
I feel like you shoudlve explained the 21:00 mark better. Professor's often time assume students just know what you're talking about
Try the new series. Maybe I explained better.