15:00 Weirdly enough, when I had my first lessons in logic classes at university, I had a better understanding of implication using ¬P ∨ Q instead of P→Q, especially the case False → False "equals" True. Anyway that Logic series brings me a lot of memories. Ok, that’s a good place to stop.
11:00 I am slightly confused by this. It states that the falsity of (P and Q) equals the falsity of P or the falsity of Q. Because of the "or" these are mutually exclusive and exhaustive of the total solutions. But there exists a third solution: not(P) + not(Q). If both were individually false, they would also be compositely false, and therefore not(P and Q) = not(p) + not(Q) . Is this correct?
I’m not exactly sure what you mean by the “+” symbol but Michael’s statements at this point in the video are following standard definitions. “or”, the V operator, is not an exclusive or: P V Q is true if either or both of P and Q are true. To look at the negations, either (P and Q) is true, or (P and Q) is false. For (P and Q) to be true, it must be the case that both P must be true and Q must be true. Therefore in any other case, (P and Q) is false: this includes the case when both P is false and Q is false, and also the cases when one of them is true and the other is false. Thus, not(P and Q) is equivalent to (not P) or (not Q). Hope that helps.
Is this the same as material implication? It seems to be equivalent to logical consequence considering in the examples the truth of a statement depends on whether or not a statement logically follows from another. However the truth table is the same as material implication...
@@h4z4rd28 I was responding to the beginning of the video where he was listing different common ways of saying the same thing, and I was asking if “P mutually implies Q” was an accepted way to say it.
proof the truth of this mathematical statemant (if you were my math teacher when i was in college) then (i would have done math for the rest of my life) Here we have P the Q always truth.
Hmm pretty interesting... So basically if we define some axioms(statements) as terminals, and define logicals , if, etc. as transitions, we would achieve a grammar, whose L(G) would be all the theorems. If so it's provable that in general it's undecidable to know if a theorem t is in L(G), or with other words that theorem is true for axiom base. A truly rough idea, but sounds interesting to investigate.
15:00
Weirdly enough, when I had my first lessons in logic classes at university, I had a better understanding of implication using ¬P ∨ Q instead of P→Q, especially the case False → False "equals" True.
Anyway that Logic series brings me a lot of memories. Ok, that’s a good place to stop.
8:20 Is this actually a biconditional statement? f'(x) = 0 could occur at an inflection point that is not a local extremum, no?
You are correct
Maybe he needs to add that f''(x) is not 0.
Maybe he meant f’ passes through 0, i.e., f’ changes signs.
As always - relaxed, down to earth and crystal clear. Great work!
Keep going with the logic videos they're great. Good job
11:00 I am slightly confused by this. It states that the falsity of (P and Q) equals the falsity of P or the falsity of Q. Because of the "or" these are mutually exclusive and exhaustive of the total solutions. But there exists a third solution: not(P) + not(Q). If both were individually false, they would also be compositely false, and therefore not(P and Q) = not(p) + not(Q) . Is this correct?
I’m not exactly sure what you mean by the “+” symbol but Michael’s statements at this point in the video are following standard definitions. “or”, the V operator, is not an exclusive or: P V Q is true if either or both of P and Q are true.
To look at the negations, either (P and Q) is true, or (P and Q) is false. For (P and Q) to be true, it must be the case that both P must be true and Q must be true. Therefore in any other case, (P and Q) is false: this includes the case when both P is false and Q is false, and also the cases when one of them is true and the other is false. Thus, not(P and Q) is equivalent to (not P) or (not Q).
Hope that helps.
P or Q in math/logic/computer is not exclusive. We specify the exclusive version with “but not both/exclusively/xor”
The second biconditional statement is false. Consider f(x) = x^3. Then f'(0) = 0, but f has no local extreme at x = 0.
Or any odd power of x really.
f'(x)=0 is a necessary condition, but f"(x)/= 0 is the sufficient condition for this case
Typo (chalko?) at 7:40: "if and only if" rather than "if and if."
Let's support his channel by not skipping the ads.
Distributive rule of OR operation over AND operation is it P v (Q^R)=(P v Q)^(P v R) ?
Yes
Prof Penn do you teach complex tetration?
need of the hour!!! Even my Math teacher writes that thing in the thumbnail wrong :'(
Is this the same as material implication? It seems to be equivalent to logical consequence considering in the examples the truth of a statement depends on whether or not a statement logically follows from another. However the truth table is the same as material implication...
could you also say something like "P mutually implies Q" as another way to say P iff Q?
Isnt that the actual definition?
@@h4z4rd28 what do you mean?
Definition of biconditional statement is that p implies q and q implies p at the same time
@@h4z4rd28 I was responding to the beginning of the video where he was listing different common ways of saying the same thing, and I was asking if “P mutually implies Q” was an accepted way to say it.
@@nathanisbored i would say yes, coz it looks like no one else is gonna respond
proof the truth of this mathematical statemant
(if you were my math teacher when i was in college) then (i would have done math for the rest of my life)
Here we have P the Q always truth.
If we negate the "exclusive or" table will be equivalent to "biconditional" table,
i want to see this in your next logical video ♡♡♡
based michael with the green new deal shirt!!
Hmm pretty interesting... So basically if we define some axioms(statements) as terminals, and define logicals , if, etc. as transitions, we would achieve a grammar, whose L(G) would be all the theorems. If so it's provable that in general it's undecidable to know if a theorem t is in L(G), or with other words that theorem is true for axiom base. A truly rough idea, but sounds interesting to investigate.
Dope
nice
5th
^^
oooh first
u should get cash app. ppl would tip a few bux