Logic - 2018-10-02

David Reed

00:01

@secret By contradiction, suppose $\pi = \frac{a}{b}$ for some $a,b \in \mathbb{Z}$. (i.e. that $\pi$ is rational. We seek a contradiction.

Recall that $\sum a_n < \infty \implies a_n \to 0\text { as }n \to \infty$, and that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ converges for all x.

So $$e^{\pi a} = \sum_{n=0}^\infty\frac{(\pi a)^n}{n!} < \infty \implies \frac{(\pi a)^n}{n!} \to 0 \text{ as }n \to \infty$$ Thus we can choose an $n \in \mathbb{N}$ sufficiently large such that $$\frac{(\pi a)^n}{n!} < 1/\pi \implies \pi \, \frac{(\pi a)^n}{n!} < 1$$

David Reed

04:32

@Secret Let me know when you want the whole thing actually. I have it typed up now but need to make sure it doesn't disappear while you are away

user21820

08:40

@user525966 The system I gave you is minimal in the sense that if you take out any rule it will be unable to prove some tautologies. ProofMood's variant for propositional logic would be minimal if you take out either ¬-Elim or ( both ⊥-Elim and LEM ).

The full system here for (many-sorted) first-order logic, specifically all the labelled rules listed under "Boolean operations" and "Quantifiers", is minimal in the same sense. ProofMood's system for (ordinary) first-order logic, on the other hand, is extremely non-minimal.

Non-minimality is frequently desirable in practice, because appropriate redundancy can greatly facilitate understanding and even greatly shorten proofs. For example, in my post I explicitly mentioned that one can have variable-renaming rules, which are redundant but can greatly shorten a proof, not to say make the core logical reasoning clear. After all, nobody wants to read a portion of a proof whose sole purpose is to prove the same thing but just with different variable names!

So it is not a surprise that ProofMood's system is quite far from minimality.

A balance must be sought for, though. A system for first-order logic with a thousand rules would be quite ridiculous and harder to use unless you already knew (in your head) a sufficiently simple system! It is just like playing a game; if it has 1000 rules and you do not know which 980 of them are redundant, it would take you a very long time to figure out how to play according to the rules.

Similarly, I can review ProofMood's system (and other ND systems) the way I am doing now, basically because I already understand a minimal system that I can compare it to. This is one key advantage of knowing at least one minimal system, even if one normally uses an extended system with extra rules for pragmatic convenience.

And I should say that it is usually easy to prove non-minimality of a given system's rules, because all one has to do is to show that at least one rule is redundant, namely to show how to convert any proof in the system into a proof that does not use that rule. In contrast, it is usually hard to prove minimality of a given system's rules, because one would have to use ad-hoc methods for each rule to show that there is some tautology that cannot be proven without that rule.

David Reed

10:02

@Secret After this one, if you want another cool number theory proof I would highly recommend erdos' proof of Bertrand's postulate. Super slick.

@user21820 hello there

user21820

Hello!

David Reed

forgot you're on the other side of the globe

user21820

Haha..

David Reed

I made the mistake of taking a stimulant a little too late ;)

user21820

How do you know I'm not right behind you? =P

David Reed

10:03

haha

user21820

Did you look?

=D

Anyway, how are things?

David Reed

how are things these days? You wrapping up on your data structures paper?

haha

not bad...im in RI taking care of my 84 yr old grandmother

occupies most of my time

I've also got a lawsuit against the state of Arizona that's going to be increasingly difficult to keep up with

Other than that its mainly MCAT review

user21820

I'm fine as usual. Hmm I suppose all lawsuits are a bother.

It seems most of them are good only for the rich.

David Reed

First two motions, although sort of trivial, were granted, which was pretty exciting

I'm asking about $75,000

I spent 100's of hours pouring through legal texts preparing for it

How much longer do you expect till you'll be Dr. user28210

FYI if you followed that big controversy with Donald Trump pardoning some Sheriff in the US...that is the defendant in this case.

user21820

@DavidReed 1 to 2 years perhaps.

@DavidReed Nope I haven't been following.

By the way, should we move this to another chat-room?

David Reed

10:19

@user21820 Ah.

https://en.wikipedia.org/wiki/Joe_Arpaio

Anyways, you seem to be busy. I mainly just stopped in to ping secret. Actually need to get back to work myself. Till next time!

Trash is fine

user21820

@DavidReed I see thanks for the link.

Secret

12:50

ok, but it is a lot less transparent compared to the e is irrational proof on what motivates us a lot of things for example:
1. The proof by contradiction obviously want to take advantage of the behaviour of $\pi$ under sine, which is why it is seeking to use $\int f(x) \sin x dx$ in one of its steps
2. Why are we choosing the nth term of $e^{\pi a}$ to establish that bound of less than one?
3. How actually is $f(x)$ motivated other than it will bring the result within [0,1] to trigger the contradiction?

Overall, there is one thing I don't understand about these more advanced irrationality proofs. Why are we looking at the nth term of an expansion and seek for a contradiction that way. Or more precisely, why the fact that these numbers are irrational will trigger the contradiction and hence motivate us to prove by contradiciton that way

Another way to ask this question is. Since I am still very terrible at dealing with negative abstractions (objects of the form there does not exist X such that P(X) is true), if this proof is straightened into a direct proof, what property between rationals and irrationals lead to the conclusions. Is it that they cannot die off fast enough in their nth term approximation, or other things?

I think usually I will be confused alot even if a proof is logically consistent when I don't understand how does a proof came to be. I am not sure if that is semantics or metalogic, but usually I will need to understand this in order to understand the proofs themselves

and usually, proof by contradiction has a lot of these auxillary elements in it that I have no idea where they came from or what reason motivates them

user21820

13:32

@Secret I think you realize by now that advanced mathematics is often ad-hoc. Proofs of irrationality are by definition proofs that the quantity in question is not rational. So it is unsurprising if they not only are not constructive but also do not seem to tell you anything more than the irrationality of that number.

David Reed

@secret hello there sir

Sometimes in math people come across things by accident

Euler discovered the gamma function by playing around with something completely unrelated, and through the most obscure and irrational substitutions and manipulations

I can't tell you what motivates the function $f(x)$ in Niven's proof

820 is correct in that the only feasible way to prove something is irrational is to show that it is not rational

Proof's by contradiction form the foundation of virtually every proof like this that I am aware

Sometimes you just have to settle with proving the result, and leave the discoverer's mindset to mystery

Secret

But it will be nicer if we can understand what aspect of irrationality we are exploiting in these proofs, otherwise I don't see how we can get very far in irrational numbers without fully understanding the entity we tried to prove outselves

David Reed

You've proven that $\pi$ cannot be expressed as a fraction of integers

that is a worthwhile result

Secret

True, but it is still deeply unsatisfying on how the inability to express as fraction of integers will trigger the contradiction. Usually in proofs, even highly nonconstructive ones, if I understand what and how the contradiction is triggered, it is more fruitful than just knowing the result of the proof

David Reed

sorry I forgot it doesn't show you that I'm still typing :)

One moment

Secret

13:43

For example, proofs of vitali sets basically exploits that these sets are not invariant under rotations and translation symmetry. That's a much cooler property than simply showing something exists

(I am typing too lol_

David Reed

Yes. So thus far we have not even finished the proof

which arguably contributes to you being confused about it

but the contradiction is effectively as follows:

Assume that pi is rational. Show that this implies that there exists an integer in the interval $(0,1)$. Note that this is not possible

There are a variety of ways to rigorously make that a statement of the form $A \land \neg A$

Let me know if and when you would like me to post it step by step and we can go through it together

Secret

I think we might need to get through it step by step. It is not immediately obvious what does irrationality has anything to do with where an interger lies in

David Reed

It doesn't

The idea to focus on is that the assumption of rationality necessarily implies something impossible

and so can't be true

user21820

@Secret If you read the proof David sketched, it relies on the fact that a certain integral involving a,b is an integer if a,b are integers, and the other properties of that integral rely on π = a/b.

David Reed

@secret Here lets go into your lab and I'll post it piece by piece

Secret

13:57

21820: that is one specific integral that so far seemed to pop out from nowhere other than by accident, thus I felt so disconnected by the flow
David: Sure, let's go there

user21820

@Secret I can't tell you what was going on in the head of the first one who discovered that integral, but it's reasonably simple that it would be eventually discovered. Also, the more advanced the mathematics the more ad-hoc the objects constructed. If you take a look at olympiad problems, you will wonder at the incredible way in which some solutions seem to fall out from thin air.

Equally incredible is that some olympiad participants can find those same solutions or alternative but equally ad-hoc solutions without having seen such problems before.

Secret

yeah, I still yet to understand the magic of number theory

user21820

@Secret For a simple example, consider the proof of Fermat's little theorem, namely x^(p−1) ≡ 1 (mod p) for any prime p and integer x such that gcd(x,p) = 1, by considering the product of { x·k : k∈[1..p−1] } mod p, and proving that this set is identical to [1..p−1] mod p, and so the extra factor of x^(p−1) must be 1 mod p.

user525966

a lot of interesting results appear to come from intuition and experimentation rather than any kind of obvious, gradual progression

come up with the cool result first, prove/ask questions later kind of deal

user21820

How is one supposed to think of that product? I don't know; I saw that trick somewhere so I never found it myself. But after you see that trick it is not so hard to extend that trick to prove Euler's phi theorem, namely x^φ(n) ≡ 1 (mod n) for any positive integer n and integer x such that gcd(x,n) = 1.

@Secret: If you try it, you may run into some issues that have to be solved that don't show up in the prime case, but at least there is no 'trickery' to extract out of thin air.

Secret

14:14

Hmm... the product seems like a brute force approach in that since we want to show $x^{p-1} \equiv 1 \mod p$ we might as well try all possible products in mod p and showing that this reduces to [1...p-1] mod p which then the conclusion 1 mod p quickly follows

not sure if I understood correctly

David Reed

Secret you should have had some exposure to group theory when taking physical chem

The above is automatic from lagrange's theorem

user21820

@Secret You need to work it out properly then you will know what is going on. Anyway my point is that it is ad-hoc. I don't think you would think of the product of { x·k : k∈[1..p−1] } very easily, if at all, if you have never seen this trick before.

Secret

right

user21820

But like Niven's proof, this construction has a low enough complexity that eventually someone will try it and find that, hey it gives a nice proof.

So I can't even tell you whether it was an accidentally found proof or not!

@DavidReed Going via Lagrange's theorem is a rather distinct route, arguably very different in algebraic structure, than this particular proof.

David Reed

I did not realize you were choosing a proof to demonstrate a more general idea

sockevalley

14:22

Good day fellow logicians, anyone here that is familiar with prolog?

Secret

I don't think lagrange theorem will be taught in point group chemistry. We only need to know basic group operations and character tables, we don't even need to know what is a subgroup

user525966

Proofs of Fermat's little theorem

This article collects together a variety of proofs of Fermat's little theorem, which states that a p ≡ a ( mod p ) {\displaystyle a^{p}\equiv a{\pmod {p}}} for every prime number p and every integer a (see modular arithmetic). == Simplifications == Some of the proofs of Fermat's little theorem given below depend on two simplifications. The first is that we may assume that a is in the range 0 ≤ a ≤ p − 1. This...

the main "trick" is realizing that the sequence a, 2a, 3a, ..., p-1 all taken mod p gives you a rearrangement of 1, 2, 3, 4, ..., (p-1), so you end up with factorials on both sides of the congruence that you can cancel out, which gives you a^(p-1) = 1 mod p

user21820

@DavidReed Yea I was. The interesting thing is that if you investigate group theory first, then you may chance upon the Lagrange theorem way before the approach I sketched above. Yet the core algebraic structure underlying Euler's phi theorem is the fact that gcd(x,n) implies existence of multiplicative inverse of x mod n, and so every proof appears to pass through that bottleneck. In particular the Lagrange theorem one seems (from this viewpoint) to have lots of unnecessary obscuring frills.

@user525966 Interesting; Fermat had the theorem in 1640, and that article seems to suggest that the first known exposition of the proof I sketched was in 1806. That's quite long for a 'simple enough' ad-hoc construction. =)

I wonder whether the related proof for Euler's theorem appeared relatively quick after that, which would support the view that it is the same 'trick'.

Secret

(Random explosive generalisation pondering to be dealt with another day: I wonder if there is a field of mathematics that tries to quantify how fast, how often and what pattern proofs in a given field of mathematics emrges, and thus forming a network of interconnecting proof ideas that feeds into many proofs)

user21820

The section mentioning applying the trick to Euler's theorem doesn't mention when it was discovered haha..

@Secret Lol... A lot of it is luck.

David Reed

14:33

@user21820 The motivation for abstract algebra is effectively that. Mathematicans discovered they were effectively proving the exact same thing over and over.

Secret

Unrelated thing about myself: I am not a very spiritual person because I am that kind of person who will refuse to use a given magic if I don't fully understood it

If someone tells me to believe in a religion, I will give them a huge pile of question marks

David Reed

I believe a scientist is necessarily an agnostic

user21820

@DavidReed For algebra, yes. But I think there is hardly any correlation between the patterns that we know of today and how fast something is found. Some theorems or proofs just escape all the mathematicians for a long time.

David Reed

That was actually the motivation for topology as well come to think of it

Secret

I am indeed an agnostic (I used to be atheist back in 2011)

user525966

14:37

i can't find anywhere talking about how euler discovered his theorem

Secret

I won't satisfy with something that is not explained, unless it is provably unexplainable like actual magic

user21820

Partly also because mathematicians are not math-machines, no matter how close the two phrases sound, so they try things more according to their interests or fancy, and not like a breadth-first-search of proofs.

user525966

most atheists are agnostic

David Reed

@user525966 there are 50 million "Euler's theorems"

Secret

As 820 knew about me some days earlier, I strongly believe in operationalism

David Reed

14:38

Are you referring to the one involving, e,i,pi,0 and 1

Secret

Operationalization

In research design, especially in psychology, social sciences, life sciences, and physics, operationalization is a process of defining the measurement of a phenomenon that is not directly measurable, though its existence is indicated by other phenomena. Operationalization is thus the process of defining a fuzzy concept so as to make it clearly distinguishable, measurable, and understandable in terms of empirical observations. In a wider sense, it refers to the process of specifying the extension of a concept—describing what is and is not an instance of that concept. For example, in medicine, the...

user525966

no that's euler's formula

David Reed

Oh you're talking about the group of units mod n

Secret

that is, anything, however confusing, will be characterised when we record how it interact with other things enough

user525966

euler's theorm is a^phi(n) = 1 mod n

Secret

14:38

and that's what I used to understand infinite sets

David Reed

Euler's opera omnia is published

Secret

This obviously failed for indefinable reals however

David Reed

You should be able to find his original publication of it translated to english

user525966

although there does seem to be another "euler's formula" in graph theory, en.wikipedia.org/wiki/Planar_graph#Euler's_formula

David Reed

@user525966 en.wikipedia.org/wiki/List_of_things_named_after_Leonhard_Euler

:)

user525966

14:40

holy

that guy must have been pretty busy

Secret

Euler–Fokker genus

In music theory and tuning, an Euler–Fokker genus (plural: genera), named after Leonhard Euler and Adriaan Fokker, is a musical scale in just intonation whose pitches can be expressed as products of some of the members of some multiset of generating prime factors. Powers of two are usually ignored, because of the way the human ear perceives octaves. An x-dimensional tone-dimension contains x factors. "An Euler-Fokker genus with two dimensions may be represented in a two-dimensional (rectangular) tone-grid, one with three dimensions in a three-dimensional (block-shaped) tone-lattice. Euler-Fokker...

lol

user21820

@user525966 He was, even after he lost his eye-sight.

David Reed

I believe he was the second most prolific mathematician of all time

First if you restrict his publications to original research

user525966

who was the first?

David Reed

erdos I think

user21820

14:42

Leonhard Euler

Leonhard Euler ( OY-lər; German: [ˈɔɪlər] ( listen); 15 April 1707 – 18 September 1783) was a Swiss mathematician, physicist, astronomer, logician and engineer, who made important and influential discoveries in many branches of mathematics, such as infinitesimal calculus and graph theory, while also making pioneering contributions to several branches such as topology and analytic number theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion of a mathematical function. He is also known for his work in mechanics...

user525966

hilbert? erdos? gauss?

i know terrence tao considered hilbert a jack of all trades

David Reed

"Erdős published around 1,500 mathematical papers during his lifetime, a figure that remains unsurpassed"

user525966

The World's Best Mathematician (*) - Numberphile

►

7:58

user21820

@user525966 Lol, please note that Numberphile is full of nonsense.

user525966

what do you mean

user21820

14:45

Not that this particular bits are nonsense, but it's too much nonsense that I actually recommend against it.

It's hard for non-mathematicians to tell which bits are nonsense.

user525966

what's an example of nonsense

Secret

the famous -1/12

user525966

the -1/12 stuff i consider nonsense if that's what you are referring to

user21820

Yup that is one.

Some parts of their videos about infinity are also rubbish.

It's unbelievable how many non-mathematicians actually buy that kind of rubbish.

But I suppose more people on Math SE know better.

user525966

it reminds me of those divide by zero proofs that show 1=2 except this time the -1/12 idea is given some credibility because it happens to show up in some string theory framework

(which imo would suggest the string theory framework is flawed but I don't know anything about it to suggest how)

Secret

14:48

blame the physicists for 1/137 and -1/12 to be made weird and arbitrary connerctions

user21820

@user525966 Even if it does show up in string theory, which anyway is not even science, it is no better than saying that 1/(1−x) = 1+x+x^2+... and hence 1/2 = 1−1+1−1+...

The flaw is obvious to anyone who knows basic real analysis; the first identity only applies for |x|<1, and if you apply it to other things like cows you will get nonsense.

user525966

how are you getting grandi's series out of that

user21820

@user525966 x = −1.

user525966

doesn't 1/(1-x) mean every exponent is 1^n?

Secret

$$\frac{1}{1-@cows} = \text{Spherical cow nonsense that constantly multiplies}$$

user525966

14:50

i think that would be 1/(1+x) would it not?

user21820

If you treat the first identity as true for x = −1, you get what I wrote, but of course it is meaningless for x = −1.

Substitute and see.

Holo

@user525966 1/2=1/(1-(-1))"=" 1−1+1−1+...

user21820

@Holo Ooh hello there!

user525966

p = 1 + x + x^2 + x^3 + ...
xp = x + x^2 + x^3 + x^4 + ...
p - xp = 1
p(1-x) = 1
p = 1/(1-x)

Holo

@user21820 Yo!

@user21820 I couldn't write down what you told me yet because real life is a bitch but I hope to get to there soon

@user525966 this is assuming p exists, which it does only if |x|<1

user21820

14:52

@user525966 Wait, be careful. You should explicitly state that x is a real or complex number and |x|<1, otherwise your very first line is meaningless. And that is the key; you can only define p to be that infinite series after you have proven that the series converges.

user525966

yeah I get the |x|<1 part

user21820

So basically anyone who is abusing the identity is ignoring the necessary condition of |x|<1 for the identity to be even meaningful.

And that is why Numberphile is not a good source to learn mathematics from.

user525966

oh i made a mental math error

ok with you now

famesyasd

do I introduce the notion of series via recursion theorem?

user21820

@Holo Sure sure; by the way I bookmarked that discussion in case you want a permalink to refer to (and also so that I can unpin from the star board haha):

Veblen Ordinals (Part 1)

Sep 29 at 8:45, 2 hours 2 minutes total – 152 messages, 3 users, 1 star

Bookmarked yesterday by user21820

user525966

14:54

you get 1/2 = 1-1+1-1+... if x = -1 which fails the condition that |x|<1

user21820

@user525966 Right.

@famesyasd You need the sequence of partial sums, so yes!

user525966

i don't think it makes sense to assign a finite value to something that doesn't converge to anything

kind of a paradox

Holo

Even the video of Numberphile that does makes sense they don't explain why it make sense

famesyasd

do I need a strong version for this?

or mine where F : A -> A and a in A

user21820

@user525966 Tell that to those people who don't know any real analysis but want to believe that 1+2+3+... is negative...

Holo

14:56

@user21820 Haha, already put a star on this so I can get the permalink any time I want :-)

user525966

ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12

►

6:18

user21820

@famesyasd Just that will do, because we are dealing with reals, not funny complicated sets. We only need to define an f∈func(nat,real) for the terms, and a g∈func(nat,real) for the partial sums, and we can do all that with the simple recursion theorem.

@Holo I have the power to unstar it haha..

user525966

also 0:43

Holo

@user21820 No(oooooooooooooo)!!!

user21820

Anyway, I'll be away. Feel free to carry on!

Holo

14:59

Bye @user21820

@user525966 this is not standard analysis

user21820

16:00

32

A: Why is this definition of convergence incorrect?

I am not sure why the other current answers are interpreting your second definition as invalid, because as a native English speaker I interpret it to mean exactly the same (logically speaking) as the first definition. In particular, you did not switch the quantifiers, even though the surface text...

Holo

This is why we should speak using only logic quantifier

MatheinBoulomenos

@Holo $\forall \exists \exists \forall \exists \forall \forall \forall \exists$

Holo

$\exists\!\!\!\forall$!!!!

$\forall^{\exists}$

user21820

16:30

@Holo Ultimately, yes. But even for logicians it can be hard to always keep the entire string of quantifiers in mind. That's why we devise definitions to condense pieces of it.

Holo

@user21820 Yes, but the english language is not made of definitions using quantifier

user21820

And it turns out that we don't need the ε-δ style of definition for real analysis. There is an alternative approach that is equivalent but has much fewer quantifiers, though it is not often taught (although Knuth has been a well-known proponent of a similar approach):

7

A: When is the derivative of an inverse function equal to the reciprocal of the derivative?

The answers so far are arguably incorrect; they merely give sufficient but not necessary conditions, and one of them even states that their conditions are necessary. We do not need differentiability in some (open) neighbourhood of the point, even for the conventional (very restrictive) definition...

The reason for the disparity is that the alternative approach is in the 'forward' direction (from δ to ε) and retains more information (the asymptotic behaviour) and hence one can easily manipulate it as needed. In contrast, the conventional definition (from ε to δ) is backward and only works when the limit exists, hence direct proofs are less re-usable.

Holo

@user21820 what other definition, in the linked answer he uses the $\frac{\Delta y}{\Delta x}$ but this is essentially the same no?

user21820

16:48

@Holo It's not really the same. In that framework you have literally variables that vary with a parameter, which of course you can capture in a first-order foundation via either sequences or functions. If we go with the function approach, for variables x,y varying with real parameter t, the notation "dy/dx" would effectively be defined as the partial-function r such that y(t+δ)/x(t+δ) ≈ r(t) as δ → 0. The quantification over ε has been 'factored out' of the symbolism, as in the intuition.

In particular, what we often have is the context "δ → 0", within which we can perform ordinary reasoning and make some additional assertions such as "δ ≠ 0" and "δ ≈ 0", and hence obtain other assertions, which we can then pull out from that context to obtain statements about limits and derivatives.

@Holo: Perhaps it would be easier to look at some examples of how this framework has much lower quantifier-depth than the conventional approach (even with limit theorems). I have some under "Asymptotic expansions" on my profile. In particular the last example links to a gigantic one:

6

A: Limit with a big exponentiation tower

Here is a very simple way using only the asymptotic expansion $e^t \in 1+t+Θ(t^2)$ as $t \to 0$ As $x \to \infty$: $\Large e^{e^{e^{x+e^{−\big(a+x+e^x+e^{e^x}\big)}}}}−e^{e^{e^x}}$ $\Large = e^{e^{e^xe^y}} − e^{e^{e^x}}$ where $\large y = e^{−e^{e^x}-e^x-x-a} \in o(1)$ $\Large \in e^{e^...

Holo

@user21820 I will look at the example

user21820

17:04

@Holo: Will be off again. See you!

Holo

See you later

David Reed

20:58

Its more of a corollary to Lagrange's Theorem. For any finite group $g^{|G|} = e$. You DO actually need basic results from modular arithmetic to get this result but they form the basis for the analysis of any finite group.

In particular you need $g^{n_1}=g^{n_2} \iff n_1 = n_2 \mod |g|$. This gives $g^n = e \iff n \big| |g|$. Its then automatic from the fact that the order of the subgroup generated by an element is equal to the order of the element and the fact that the order of any subgroup divides the order of the group (lagrange).

I do agree though that there is value in learning the old-fashioned way.

Actually I've frequently found in mathematics that I'll spend a whole page proving a result only to develop the machinery a few semesters later to prove it in a single sentence.

$|g| \bigg| n$*

Veblen Ordinals (Part 1)

Transcript for

♦ $Mathematics$ Logic

Veblen Ordinals (Part 1)

Transcript for

♦ Logic

♦ $Mathematics$ Logic