Discrete Math 1.3 Propositional Equivalences

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[tariqelamin]

Years later and your vides are still valid, honestly you are a life saver

[TheFhdude]

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)

[maisamnoyan]

Thank you so much for the easy explanation of lectures. It was so helpful for me and I appreciate your work.
Good Luck

[reasetilo]

15:06
You just dropped the addition sign, why?

[cesar]

11:35 Should be !(P && Q) -> (!P OR !Q)

[abdulrahmanal-eryani]

Thank you so much for your nice and easy explanations

[Boundlxss__]

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?

[bradley]

What's the name of the rule used at 20:59?

[kevinz]

At 27:20, for problem #3, wouldn't the solution be satisfied if p = T; q = T; r = F?

[kayd]

I feel like you shoudlve explained the 21:00 mark better. Professor's often time assume students just know what you're talking about

[dzapwince]

I think for no 3 if you make P = T, Q = F, R = F. it would be satisfiable