Hey love your channel and may I ask a question: If in set theory, I can create a relation which takes a set of elements which are propositions (like set a is a subset of set b) and map it to a set of elements containing “true” and “false”, then why is it said that set theory itself can’t make truth valuations? I ask this because somebody told me recently that “set theory cannot make true valuations” Is this because I cannot do what I say above? Or because truth valuations happen via deductive systems and not by say first order set theory ?
Would some smart person care to clear up something very basic for me: I could easily follow everything Trev was doing here. However ... I learned that |T->T|=T; |F->T|=T; |F->F|=T; |T->F|=F. And that even makes intuitive sense to me. If we have p->q, it means that p "guarantees" q. So if p and q are True, then the whole statement is True, because p did "guarantee" q. If we have that p is True and q is False, then the statement is false, because p did not "guarantee" q, which was the condition. F->F and F->T being both True also make sense, because it is possible that if there was no p, then there was no q, and it is also possible that even without a p there was a q. But here, in the 3rd example for instance, we said alright, we want to check whether or not it is the case (is it True) that if we have "A^B", then we have "C", assuming that everything to the left is True. So, we assume that "A^B" is indeed true, and see what happens, i.e. we assume that "A^B" is true, and check whether "C" is true. If it is, then we have proven that "A^B" gives us "C", when the stuff to the left is true. This too makes perfect intuitive sense, as much as the previous stuff. But what I don't get is ... "(A^B) -> C" is true even if "A^B" is false !!! Again, assuming here that "A^B" is true makes total sense, it even feels commonsensical, because we want to check IF "A^B" is true, is "C" then true, which it is. But ... if A is true and B and C are false, for example, then we get that "A->(B^C)" is False, and that "(A^B)->C" is true! So, what if we had shown that? We wouldn't then have known if the right-hand side follows from the left-hand side. I know that I'm missing very basic stuff here. But my main point is that, even though all of this makes complete sense when I accept the defintions, a lot of this stuff seems random. Sometimes F->T is true, and sometimes p->q means that if p is true then q is true (well, it always means that, but...), so we assume that p is true and check if q is true, and if it is, then we know that p->q is true... but isn't that just one case of p->q being true? T->T. Why are we only interested in that? Why not in F->F and F->T? It doesn't seem as if the definitions ever really change, but they seem situational somehow, as if we use one part of the definition for one kind of problem and another for another kind of problem.
You say: "But what I don't get is ... "(A^B) -> C" is true even if "A^B" is false !!!" The only way you might get surprised of this is if you did not accept the fact that from something false, nonsensical, we could get any conclusion. In fact from (A^B) false you get both (A^B) -> C and (A^B) -> not(C) Which would both be true at the same time I'd say the point of this is so that we can talk about the case where (A^B) is true, and say "and whatever if it's false, would be nonsensical anyway" If you instead wanted to talk about what happens when (A^B) is false, you would use not(A^B) (A^B) -> C means: if (A^B) then c, else i'm not saying anything not(A^B) -> C means: if not(A^B) then c, else i'm not saying anything When we are not saying anything (case F->T or case F>F) we just leave the statement true because it's not saying anything anyway (besides that anything follows from nonsense) and because it's more practical to have it set up this way
Are rules the same as theorems? If there were all the theorems we would have all the results. So, if we'd have one of those theorems would we imply the other ones with it?
We use the rules to prove theorems. Once we have proven a theorem (like commutativity, DeMorgan's, Modus Tollens) then we can use it whenever. Different systems start out with a different set of rules.
Forget the Greeks. Just forget them. They had a good run. It was great. But just let them go now. Please. If we can drop them from keyboards, surely we can drop them from everywhere else. Also, where can I get me a 4th century Greek keyboard?
This was a great video. Very clear and easy to understand because of amazing handwriting and very good speech skills. Thank you for this!
I cannot describe enough how much of a lifesaver your videos have been for my class. Thank you so much
hangi okul kanka :D
@@Po1SoN333 ali adı başka ülkelerde de kullanılıyor. profiline bak zaten.
@@Ali-kh1rs which subject ur school contain this
Amazing video. Thumbs up all the way. Thanks for making this content available online. It makes a big difference!
Do we really need line 3 to exist? (referring to 19:58 example # 3), in my answer, i went to line 4 straight and reached the same conclusion.
Very well explained. Cleared my mind about the topic thank you very much bro
man tysm this saved me a lot of time
Great video, thank you!
is reiteration or hypothesis the equivalent to 'assumption' which is what other people use in natural deduction
Hey love your channel and may I ask a question:
If in set theory, I can create a relation which takes a set of elements which are propositions (like set a is a subset of set b) and map it to a set of elements containing “true” and “false”, then why is it said that set theory itself can’t make truth valuations?
I ask this because somebody told me recently that “set theory cannot make true valuations” Is this because I cannot do what I say above? Or because truth valuations happen via deductive systems and not by say first order set theory ?
Thank you very much for this video
19:25 if we Took A as hypothesis still we will be able to prove it.
Thank you so much!!
Great , thanks man for that
How many lines should be cited for an application of &I?
Which is best out of truth tables, tress and natural deductive logic? When would you use each? Or do you just use one.
Would some smart person care to clear up something very basic for me:
I could easily follow everything Trev was doing here. However ...
I learned that |T->T|=T; |F->T|=T; |F->F|=T; |T->F|=F. And that even makes intuitive sense to me. If we have p->q, it means that p "guarantees" q. So if p and q are True, then the whole statement is True, because p did "guarantee" q. If we have that p is True and q is False, then the statement is false, because p did not "guarantee" q, which was the condition. F->F and F->T being both True also make sense, because it is possible that if there was no p, then there was no q, and it is also possible that even without a p there was a q.
But here, in the 3rd example for instance, we said alright, we want to check whether or not it is the case (is it True) that if we have "A^B", then we have "C", assuming that everything to the left is True. So, we assume that "A^B" is indeed true, and see what happens, i.e. we assume that "A^B" is true, and check whether "C" is true. If it is, then we have proven that "A^B" gives us "C", when the stuff to the left is true.
This too makes perfect intuitive sense, as much as the previous stuff.
But what I don't get is ... "(A^B) -> C" is true even if "A^B" is false !!!
Again, assuming here that "A^B" is true makes total sense, it even feels commonsensical, because we want to check IF "A^B" is true, is "C" then true, which it is.
But ... if A is true and B and C are false, for example, then we get that "A->(B^C)" is False, and that "(A^B)->C" is true! So, what if we had shown that? We wouldn't then have known if the right-hand side follows from the left-hand side.
I know that I'm missing very basic stuff here.
But my main point is that, even though all of this makes complete sense when I accept the defintions, a lot of this stuff seems random. Sometimes F->T is true, and sometimes p->q means that if p is true then q is true (well, it always means that, but...), so we assume that p is true and check if q is true, and if it is, then we know that p->q is true... but isn't that just one case of p->q being true? T->T. Why are we only interested in that? Why not in F->F and F->T?
It doesn't seem as if the definitions ever really change, but they seem situational somehow, as if we use one part of the definition for one kind of problem and another for another kind of problem.
You say:
"But what I don't get is ... "(A^B) -> C" is true even if "A^B" is false !!!"
The only way you might get surprised of this is if you did not accept the fact that from something false, nonsensical, we could get any conclusion.
In fact from (A^B) false you get both (A^B) -> C and (A^B) -> not(C)
Which would both be true at the same time
I'd say the point of this is so that we can talk about the case where (A^B) is true, and say "and whatever if it's false, would be nonsensical anyway"
If you instead wanted to talk about what happens when (A^B) is false, you would use not(A^B)
(A^B) -> C means: if (A^B) then c, else i'm not saying anything
not(A^B) -> C means: if not(A^B) then c, else i'm not saying anything
When we are not saying anything (case F->T or case F>F) we just leave the statement true because it's not saying anything anyway (besides that anything follows from nonsense) and because it's more practical to have it set up this way
thank you!
Are rules the same as theorems? If there were all the theorems we would have all the results. So, if we'd have one of those theorems would we imply the other ones with it?
We use the rules to prove theorems. Once we have proven a theorem (like commutativity, DeMorgan's, Modus Tollens) then we can use it whenever. Different systems start out with a different set of rules.
Thank you! :)
In terms of the homework problems, I'm confused by what the symbol "v" stands for between A and B
“Or”
super
Not first.
Forget the Greeks. Just forget them. They had a good run. It was great. But just let them go now. Please. If we can drop them from keyboards, surely we can drop them from everywhere else. Also, where can I get me a 4th century Greek keyboard?
second!
first!