@F.Zer Indeed. A 5-cycle satisfies the conditions but has no triangle.

@ConGovDeIn "for all v and which" is STILL gibberish. I don't even know what you want to say.

@user21820 Don't worry, I got what I needed from someone else. My problem was solved.

@ConGovDeIn Of course you'll get answers from other people. I don't teach those who don't wish to learn properly.

Was your ego got hurt because someone else helped me?

Sorry. Next time I shall just ask you.

@ConGovDeIn No it has nothing to do with my ego. I stated a fact. You did not put in effort into your attempts here, and so I was simply unable to guide you to get the answer by proper reasoning.

I don't just tell people the answer (unlike others) because it won't help you learn properly.

It's not just for you; I teach everyone the same way.

To put it another way, I don't agree with the common teaching style of other people who tend to just give the answer to a question. Just looking at an answer will make you lose the opportunity of figuring out how to get it.

Yeah, I agree with that.

@ConGovDeIn: If you still want to learn properly regarding that question, you should tell me why "for all v and which" does not make sense.

Learning is not only about learning what is correct and why it is correct but also learning about what is wrong and why it is wrong.

That's what I want you (and all other students) to understand.

Actually I was on mobile and I wanted to write "for all v which are greater than 2" and I think I would have mistyped "all" so the auto correct feature made it "and".

Ahhh... no wonder...

I appreciate your philosophy/ideology of teaching.

Your phrase "for all v which are greater than 2" makes complete sense.

However, do you understand why "∀ v = S(S(0))·u" does not make sense?

@user21820 Yes, because "=" is a function and it should have "terms" on both sides of it but $\forall v$ is not a term.

@ConGovDeIn That's right.

but then what exactly is "$\forall v$"? It's not a term, it's not a formula, then what it is?

@ConGovDeIn Your "remedied" version is still incorrect, and I'm uncertain what you were trying to say.

@ConGovDeIn It's called a quantifier. You can only form a formula of the form "∀v ( Q )" when v is a variable and Q is well-defined given v.

This syntax rule already excludes your last attempt from being correct. Can you understand why?

@user21820 the things on the left of "$\implies$" must be having some "illness", eh?

@ConGovDeIn Yes. You cannot just join together two formulae, because that doesn't make a new formula.

Also, even if you insert a "∧" between them, the second piece "(∃u)(v=S(S(0))⋅u)" is still wrong according to the same rule, because v is no longer defined outside of "∀v ( ... )".

let me try:

By itself it is a valid formula. Though it is false. Do you get why?

@user21820 It says that all v's are greater than 2, but 0 and 1 are not.

Exactly.

@ConGovDeIn I think you have a fundamental misconception with how to use quantifiers. Your last attempt is better but still wrong for the same reason.

You want to say "for all v which are greater than 2, ...". I want you to finish that sentence first and then transform it (still in English) to a sentence that starts "for every v, ...".

I have a confusion: does $\exists$ mean "if there exists a ..." or does it assert that "there exists a ..."?

@ConGovDeIn The point is that every quantifier comes with a scope:

That's why I want you to rephrase in English first, to have a more intuitive grasp of "scope".

In symbols the scope is denoted by the brackets, but you're not going to understand it if you don't understand the intuitive notion.

@ConGovDeIn Neither of these are sentences...

I'm not getting the point.

"I'm not getting" is not a sentence.

What you just wrote are all not sentences.

FOL formulae correspond to complete sentences.

"for all butterflies" is not a sentence either.

"every butterfly has wings" is a sentence.

yes, yes I'm getting it a little.

"for every v, there exists a u such that v = u \cdot 2 and 2 \lt v"

Ok that is a good sentence now. Is it true or false?

false. There exist odd numbers also.

So what did you actually want to write?

> we want to write the Goldbach Theorem in logical form

every even number greater than 2 can be written as a sum of two primes.

Yes, so...

> I want you to ... transform it (still in English) to a sentence that starts "for every v, ...".

And, as I said before, you should not use inaccurate phrasing like "can be written as".

"for every v, such that v is even and greater than 2 there exists p and q such that v =p+q and p and q are prime"

@ConGovDeIn Sorry I was away just now, and may reply intermittently later as well. Your last attempt is better, but still not in the form I wanted you to write it in. In fact, it's still slightly ungrammatical.

The following is grammatical:

> For every v such that v is even and greater than 2, there exists p and q such that v =p+q and p and q are prime.

Your version has the comma in the wrong place. This sentence states something about every ( v such that v is even and greater than 2 ). What you need to rewrite it into is a sentence about every v.

You can do so using "if ... then ..."

> For every v, if v is ... then there exists ...

Finish the sentence. Then we can proceed to translating to FOL.

@user21820 why we use right hand rule to find direction of cross product?

@Rover It's definition of cross product, by convention. You can very well define it using left-handed rule.

It's the same with many other things where we need to arbitrarily pick a fixed direction to measure against, like signed angles, signed area, signed volume...

I mean how we came to know what direction will it have and why the resultant direction of vector will be perpendicular to plane containing the two vectors..

What is your definition of "cross product"?

Cross product of two vectors $\ vec{a},\vec{b}$ is $\vec{a}×\vec{b}=|a||b|sin($\thetha $), where $\ thetha $ is angle between the two vectors.

And direction is given by right hand rule

@Rover Then why do you ask "how we came to know"? Wasn't my point that it is by definition?

Yes, I am a asking how we came to this definition ?

Well, because it has nice properties.

It was used since 1773 and then in 1861 by Maxwell for his equations.

That's what I said.

Obviously, you cannot always make arbitrary choices. Sometimes, one choice forces you to choose a certain way for other things.

If you choose right-handed 3d coordinates and right-handed cross-product, then you have a certain coordinate-based formula. Switching each of those choices will change the sign of the formula.

So you can choose any two of these arbitrarily but not the third.

@user21820 Ok , so since there are two possibilities in cross product $\vec{a}×\vec{b}$ and $\vec{b}×\vec{a}$ we chose the normal to plane as they are also two ?

@Rover No, that 'reasoning' makes no sense.

@user21820 okay.

You're going to have to precisely define cross-product before you can even talk meaningfully about why it is defined that way.

The fact that you say "two possibilities in a×b and b×a" shows that you don't really know a precise definition. After all, v×0 = 0×v.

Okay, my book didn't give any precise definition .

I don't think that book is a good one. That page is not very accurate or meaningful except for the fact that the choice is arbitrary.

Okay, can you suggest any book to look for vectors and 3D geometry?

I don't have a book suggestion, but you can define it as wikipedia does.

Ok

Similar to what you wrote earlier.

The point is that you need to be precise about what right-handed rule means, including how to handle zero vectors.

It turns out to be equivalent to the formula as stated by wikipedia as well.

Ok, are wikipedia's all definition precise for maths point of view to follow?

Wikipedia tends to be precise enough for mathematics. However, sadly, it can be overwhelming in terms of content.

Ok.

Most of the time, you can get what you want by reading carefully and skipping the bits that you don't understand. In this case, you might get:

> The cross product of two vectors a and b is defined only in three-dimensional space and is denoted by a × b.

> The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

> The cross product is defined by the formula a × b = ‖ a ‖ ‖ b ‖ sin( θ ) n where θ is the angle between a and b in the plane containing them (hence, it is between 0° and 180°), ‖a‖ and ‖b‖ are the magnitudes of vectors a and b, and n is a unit vector perpendicular to the plane containing a and b, in the direction given by the right-hand rule (illustrated).

> If the vectors a and b are parallel (i.e., the angle θ between them is either 0° or 180°), by the above formula, the cross product of a and b is the zero vector 0.

Seems good enough to me.

Ok, Can I ask maths questions here ? Though they are not of logic..

This room is for all basic mathematics, which includes basic logic, as stated in the room description.

Ok

@ConGovDeIn Good, except you're missing something after "greater than".

Now it is easy to translate to FOL:

> ∀v ( v is even ∧ v greater than ... ⇒ ∃p,q ( v = p+q ∧ Prime(p) ∧ Prime(q) ) ).

Carefully note the position of the brackets, because you had a fundamental misconception with them earlier.

Also note that we adopt precedence rules for the boolean operations, highest to lowest: ¬,∧,∨,⇒. That is why we can write "A ∧ B ⇒ C" to mean "( A ∧ B ) ⇒ C".

@user21820 Thank you. Is it possible to prove 6 vertices theorem ? Isn’t a 6-cycle a counter-example ?

@F.Zer: Did you manage to prove that 6 vertices must have a triangle? And have you also started on the PA exercises? No hurry of course, but I'm just wondering what you're trying now. =)

@F.Zer Almost but in a 6-cycle you can find 3 vertices that fail to satisfy one of the conditions.

Did we just write at exactly the same time? That is surprising :-)

@F.Zer Yea... 3s off according to the time-stamps.

Wow

@user21820 Oh, I will try to find those 3 vertices that fail to satisfy one of the conditions, prove the theorem and start with PA exercises. Seems like that is the way to go :-)

Ok!

@user21820 For example: ¬i~m and ¬m~k and ¬k~i. So, that condition doesn’t hold for every pair of vertices in G.

Do you think it is correct ?

@F.Zer What is your 6-cycle?

Sorry, I will type it below:

Just list the 6 vertices in order around the cycle, and the 3 that fail the condition.

@user21820 i,j,k,l,m,n,i. The 3 vertices failing the condition are: i, m and k.

Yea then that's right!

Good !

The 6-vertex theorem is definitely correct; in the graph formulation it is usually given as an introductory problem to combinatorial graph theory.

That's why I am surprised I managed to get it wrong when translating. Come on, how can 6 become 5??

Clearly 2·3 = 2+3... heh..

@user21820 That's a good joke :-)

@user21820 $$ \forall v ( (\exists u) (v = S(S(0)) \cdot u) \land S(S(0)) \lt v \implies (\exists p)(\exists q) (v =p+q \land Prime(p) \land Prime(q) ) )$$

Am I fine this time?

@ConGovDeIn Excellent.

I'll just point out that you don't need to put so many brackets. In modern logic it is conventional to put just the brackets around the scope of the quantifier, and also to drop them for consecutive quantifiers. Some people put brackets around the quantifiers as well, but it looks more cluttered because you often still need the scoping brackets.

Can you clear one more doubt of mine? When I wrote “$ \forall v (\exists u)( v = S(S(0))\cdot u)$” how did I ensure that it meant “for all v if there exists a u such that v = 2 \cdot u” and not “for all v there exists a u such that v = 2 \cdot u”?

I mean was that “if” assumed when I used the $\implies$ arrow?

@ConGovDeIn What you wrote means the second. I do not understand why you think it meant the first. I don't see any "⇒" either.

That is why your earlier attempts were really off, and hence I wanted you to get it right in natural language (English) first.

Logic is not about symbols, but about the thinking. The point about a quantifier is that it has a scope. "∀v ( ... )" says something about every v, and that something is specified by the entire (bracketed) scope for the quantifier.

In English, we do not have brackets, but you at least have a reasonable intuition about scope of a quantifier (every/some).

Okay. Means whatever come after $\forall v$ should be a valid formula and should be proper “bracketed” and it is called scope.

In English, "for every butterfly B if there exists ..." is not even grammatical.

@ConGovDeIn Correct!

Are you available in some other general chat rooms also?

@user21820, can I interrupt with a question ?

@F.Zer Sure.

Given a simple undirected graph G with at least 6 distinct vertices such that for every vertices x, y, z in G we have x ~ y or y ~ z or z ~ x (where "x~y" denotes "there is an edge between x,y"):
	Let i,j,k,l,m,n be 6 distinct vertices of G.
	If ¬i~k:
		If ~k~m:
			If ¬m~i:
				¬i~k ∧ ¬k~m ∧ ¬m~i
				¬(i~k ⋁ k~m ⋁ m~i)
				i~k ⋁ k~m ⋁ m~i
				⊥
			m~i
		If k~m:
			...

@ConGovDeIn I maintain a few chat-rooms. This one is for basic mathematics. Another one is for (advanced) logic. Another is for computability theory.

@user21820 This is my attempt using some of the intuition gained into the problem

@F.Zer Well okay.. But the point of using the graph interpretation is to use your intuition, so you should have just written:

If not i~k and not k~m:
  m~i.
If not i~k and k~m:
  ...

@user21820 That's amazing. Good to see other ways of reasoning.

Write less, think more. When you're done then fill in the mechanical parts.

In fact, if you are clever you will design a nice way of illustrating the reasoning on the graph itself instead of writing it in if-else structures.

Good. I will try :-)

The point is not to be formal, until you believe you have a proof. =)

That last comment should be starred, I think :-)

@F.Zer Hah. If too many comments are starred, the really good ones can't be seen anymore. =P

The starboard does not allow me to pin comments that I like for more than 2 weeks.

So new starred tend to push old starred ones down into oblivion.

@user21820 That's why I hesitated to star it. If everything is important, nothing is important. The "over-starring problem", I think :-)

I guess I am starting to talk nonsense. Ignore me.

Even the experts (as you are) could have some little confusion. Don't worry :-)

@F.Zer: Ok I'll be off! See you next time!

@user21820 See you next time !