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12:08 AM
silence isn't empty - it's full of answers
@DanDonnelly If the Christians are to be believed, he did die, n'est-ce pas?
@DanDonnelly Example? Why should this have a name?
Yo, but I was speaking in the sense that as if he died very recently. In English that's called a simile.
You got schooled, son :)
But he's dead, right?
Similes are usually identified by containing like or as
Did he die of COVID? Like Emma?
12:14 AM
Okay, Xander. Why should that have a name? Because if $(n \gt 3) \wedge (3 \gt 2)$ are two propositions, then the only thing you'd ever conclude from those two, assuming that $n, 3, 2, \lt$ are defined in the logical context, would be $(n \gt 2)$
It should have a name sort of like "canonical conclusion"
There's your example and intuitively a reason would be for the sake of the computers having to do long proof-theoretic / type-theoretic computations.
Or searches in the space of proofs
I am not at all convinced that this should have a name---how often does it really come up?
Always! That's the key part that I want to prove
why can we not conclude $n>1$
If not always, then it's certainly nearly always
You have used a lot of very strong statements: "The only thing...", such things come up "always", etc.
12:20 AM
You can't because we make a definition suitable for us, and that would mean $(2 \gt 1)$ is not in the logical context, nor in our given list of input props
I am not convinced.
that message was for
k edited last message
You're not allowed to add to the context, it's held fixed throughout this problem.
The next step of course in a search algorithm would then be to add the conclusion to the context (if the algorithm did things that way)
So in axiomatic theory, you usually don't have $2 \gt 1$ as an axiom
But you'd have stuff like $a \gt b \gt c \implies a \gt c$ for the reals
So the context would act on the two inputs and conclude, but it needs some more input propositions in order to conclude @shintuku 's comment
I at least think this warrants further investigation on my part
this might have some applicability in reasoning specific to statements in one or two variables about a linear ordering. i don't see it having general applicability to 'logic'
user21820 over at logic chatroom might have an idea
Since I know a little logic / type theory, and I cannot answer this question
@leslietownes not true entirely, take an example that always comes to my mind
Take $i,j : \text{Id}_X$ i.e. the identity type on cat object $X$. Then conclude $i = j$. There is nothing else new that can be concluded.
12:26 AM
one of the confusing things about math for a lot of people is that many textbook statements, including statements in the middle of proofs, are often not the optimal deduction from what is assumed. they are shaped by what is known and what is easy to prove from what is known, which might not be the logical limit.
yeah it makes sense. maybe not in the form of conclusions, but in the form of valid semantical statements
anyway for a short answer to your question i don't know that this concept has a name. i think xander's comments and my own put pressure on whether this is a well defined concept at all. it seems to me that in some restricted universe, short of all logic, it could be well defined.
you have a set of semantic units and you figure out what are all the possible combinations of them
Sure @shintuku but this method implies an "algebra" of propositions meant to make what you say (that computational search problem) a lot quicker
$$\sum_{k=2}^{9} = \frac{1}{2}(9)(10) - 1 = 44, $$ right?
Why do I have an extra candle?
12:28 AM
It's because extra candles are useful in case you lose all the main ones
i think this could be a well defined concept, ask user21820 in the logic chatroom
when you build a first order system of logic, you define a symbol set and a syntax on it, and these limit the possible conclusions when you associate the semantics to them
Oh! I didn't read the box correctly. In big print, it says "Enough for all 8 nights of Channukah", but elsewhere on the box, it says "45 candles". They gave me an extra. Hrm.
you could probably speak of extensions in this sense
12:31 AM
Okay, I'll wait for them their, like a panther
Can we not make this discussion about religion, @Xander, @Dan? Please?
you study these extensions notably to determine completeness, consistency and other such properties
the idea you're speaking about isn't that far away from that
@amWhy sure thing, no problem!
@DanDonnelly Thanks! :-)
12:32 AM
Sry, bout that, I slipped up
@DanDonnelly No problem! You're the least of my concerns here.
@amWhy may I ask, what concerns you here?
Are you an agent of Stack Exchange?
lol :)
probably a spy hunter
@DanDonnelly Not at all, not in any way. I'm an average user. I'm from the planet Zolithia, entrusted to better understand earthlings! ;-). No, just a meager user on this site. lol :) You did nothing wrong. I just stated my request, and not just to you, mind you. But, hey, seriously, you're good to go.
@DanDonnelly here's an example, where A1-5 are rules for semantically valid constructions
I trust that you're not a double agent from Zolithia sent here to teach us mathematics, but cleverly hide the anti-gravity math, so that we're stuck on Earth.
12:38 AM
@shintuku I will brb, and I will examine that figure
auf = off, mit = with?
it's just to give you an idea about building statements in semantics there's not much to analyze
auf means on
Tell that to the Lean developers lol, proof assistant code is on extreme side of difficult
it's from pages 15-16 of Ebbinghaus' introduction to logic
But probably not more difficult than an informal proof (eventually)
otoh FLT
was 200+ pages of alien-level math
Converting that to code is probably not going to happen in the next 10 years
12:43 AM
@XanderHenderson in case one breaks!
I’m guessing packaging is more stable with a multiple of 5?
It's because primes
If you cannot divide the candles, you need to buy extra packs = more profit
I give 2 candles to you, you, and you, because we share things equally. That was their math model lol
@geocalc33 what are you doing?
Sitting on a chair
1 hour later…
2:10 AM
Ok, now I'm trying to show that $(\infty, 0]$ and $(0, 4]$ are numerically equivalent. Just from looking at them, I'm pretty sure both of these sets form a bijection with $\mathbb R$. So I will just show that, and thus they will be equivalent (numerically). I can't think of a simpler way.
yeah, they are. if you play your cards right you can even find explicit formulas for your bijections with R and compose them to get an explicit formula. but that approach sounds solid.
Oooh, I will try that.
$1/x$ gives a bijection between $[1, \infty)$ and $(0,1]$. variations on that theme, perhaps.
people goof on the few weeks of precalculus, or whatever class it is, where the study is on how various operations transform domains and ranges of functions, and how to picture them. but the usefulness of that information never really goes away.
that might be the most useful and most underappreciated aspect of precalculus, or whatever class that was. i don't think i saw it before then.
Definitely guilty of that. I blame being allowed to use a graphing calc for all of the assignments/tests, lol.
@leslietownes Yeah, it's precalc. We did a refresher week at the beginning of my calc course, and it was basically this.
Not these proofs, to be clear, but understanding how graphs work and their transformations.
i'll begin to play the game. 4/x happens to implement a bijection between $[1, \infty)$ and $(0, 4]$. 4/(x+1) happens to implement a bijection between $[0, \infty)$ and $(0, 4]$. i haven't proved these statements but they are easy to verify and we are one transformation away from getting that idea to work.
$e^x$ implements a bijection between $(-\infty, 0]$ and $(0, 1]$. a wealth of opportunities there. although depending on the class, maybe nobody knows what $e^x$ is yet.
2:22 AM
What do you mean by that statement about $e^x$?
And, wow, that was fast.
i mean that if you restrict $e^x$, sometimes thought of as a function on all of $\mathbb{R}$, to being regarded just as a function on $(-\infty, 0]$, its image is $(0,1]$, and it is a bijection between those two sets.
but proving that might be harder than proving similar properties of another function. the earlier example hints at rational functions (quotients of polynomials) being able to get the job done.
whats up everyone
Hm, ok
allie we were just discussing the consequences of the linear independence of A v_1, ..., A v_n. great that you could join us.
idk man
i drank a lot of poison tonight
jjust wanted to say hi
2:27 AM
well, drink some water before bed time.
am doing so
2:39 AM
Every countable mode of PA is isomorphic, as a semiring, to a subring of $\Bbb C$
or so they would have us believe
2:56 AM
@AkivaWeinberger isomorphic as a semiring implies isomorphism as a ring, they're the same conditions
3:06 AM
If $f : S \simeq R$, from semiring $S$ to ring $R$
Then defining $-s := f^{-1}(-f(s))$ gives you + inverses in $S$
So the natural conclusion is that every countable mode of a PA has a ring structure isomorphic to a subring of $\Bbb{C}$.
Just sayin, we don't speak in the language of monoids when things are forming groups, usually
I'm not certain you can do the same construction if you didn't have isomorphism
4:03 AM
hjey tedd
@DanDonnelly $(\Bbb N,+,\cdot)$ is a model of PA but it's not a ring
A model of PA, by definition, doesn't have an element representing -1, because one of the axioms of PA is $\lnot\exists n,S(n)=0$
4:20 AM
or so they would have us believe
poison sounds good right now
So here's a neat trick
PA is incomplete, famously; it doesn't prove Con(PA) (the assertion that PA is consistent).
I want to add an axiom to PA that convinces it that PA+Con(PA), PA+Con(PA)+Con(PA+Con(PA)), etc., are all consistent.
In fact, I want to go further; I want to add an axiom to PA so that the result knows it's all consistent.
So I want a statement x that asserts "PA + x is consistent."
What goes wrong?
caring about logic in the first place?
At first I thought it was the self-reference; the fact that x references itself
but no, there are standard techniques to get rid of the self-reference. So x actually can be written in the language of PA
4:35 AM
i went to grad school with some logic people. they were all weird. one of them wore a cape. i'm not kidding, an honest-to-goodness cape.
The issue is, PA proves that x is false
Said another way, PA+x is inconsistent (despite x claiming exactly otherwise)
@leslietownes Are you sure it wasn't a sarong
no, it was a cape. like the count on sesame street.
i remember something with sarongs but that was different.
5:23 AM
who started that sarong thing, anyhow.
sometimes after a few pints with my indo friends we start to sing a sarong or two...
then the shadow puppets come out
i would like to go back to indo. except the last time i had to give a $20 bribe to get in.
i haven't been out of the country since... hm. 2006.
oh, no, i went to canada in 2007 but canada doesn't count.
i have only been to ireland. except for barcelona & london.
well, 2006 is a long time, maybe mx as well.
5:33 AM
my wife and i went to london and edinburgh in the winter of 05-06 and then my dad and i visited london again in the summer.
my wife is heading off to hawaii for a week and a bit tomorrow. the sucker is driving them to the oak at 4:30am.
i have a tradeshow supposedly tmr
in moscone.
we went up to cambridge for a day to visit a math friend. i went through one of the buildings and changed the signs on a few chalkboards where people had written DO NOT ERASE.
i have no friends.
we used to write do not erase on blackboards just for fun
i don't like moscone very much. there's a very first-dotcom-boom aspect to all of that development. the beginning of the end.
my contract may end at eoy, and i have no leads. most of my cohort are comfortably retired with their many millions, so i can't ding them.
5:36 AM
do you have any patents? we could sue people if you had some patents. you need to still own them. it's no good if someone else has them.
my daughter is getting uk offers but she wants a us offer but has one so far.
half the family is looking for jobs
i have a few, but i gave up the rights a long time ago
they were just to keep the vcs happy. they served their purpose
my sister works mostly in restaurants, she was out of work for most of 2020 and the start of 21. she has a few jobs now. there will always be restaurants.
now cadence owns all of my patents
i am slowing down which id depressing.
my recommendation would be to avoid aging if possible.
SYSTEM CHIP SYNTHESIS. looks good although it may have expired by now.
nobody ever litigated this. this saddens me.
that was a bs one if ever there was one
5:39 AM
i wasn't going to say anything.
haha you are a patent attorney's worst nightmare. if we were suing on this, the client would retain you as a 'consultant' just to keep you from saying that.
i think that was the first, i have no idea why they added me to it other than a few paragraphs i supplied
yeah, you wouldn't be saying any of this.
you could name your price
there is a much longer and truly boring backstory to that one
it was what drove me to starting my own company.
i should go to sleep, up in a couple of hours/
vector [a,b,c,d]->[b,c,d,a] is a linear transformation right? I can't seem to remember enough to google what I need to remember how to make a matrix for this
5:45 AM
oh is this going to be the elements along of the diagonals jsut above or below the main one?
0 everywhere else and then 1 there?
$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}$
ahhh where did my mathjax bookmark go >:l
there oughta be a link at or near upper right. unless you're on mobile, in which case, heaven help you.
I realized that there is "undo" option in iphone, which works if you shake the phone a little to left and right in case you want to undo a message accidentally deleted. :)
5:50 AM
@leslietownes got it :)
6:11 AM
How big can the error be: Instead of using the binomial coefficient to calc probability of something happening over x tries, simply multiply by tries. Of course with reason, as it is obvious that the chance that I throw heads in 2 tries is not 100% ;)
probabilities like 1/1,000,000 per try
probability of hitting at least once in n tries with p probability of success is 1 - (1-p)^n. this is np + higher order terms in p, with coefficients that can grow large in front of them, and with the number of terms also growing with n.
if p is tiny and n is not too big, maybe not that different from np.
and probability per try times tries is simply the expected value or?
i wasn't computing any expected value. just the probability of, if you like, flipping at least one heads in n independent tosses, if the coin flips heads with probability p.
Hello all
Can someone just help me on how to find the integral of a unit ramp signal from t=0s to t=1s whose frequency is 1kHz?
A: Calculate the integral of a saw tooth wave

JanWell, we know that: $$\text{C}=\frac{\text{Q}}{\text{V}}\tag1$$ We also know that the voltage and charge are dependent on time, so: $$\text{Q}\left(t\right)=\text{C}\cdot\text{V}\left(t\right)\tag2$$ And, we know that: $$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\text{V}_\text{C}'\left(t\right...

The above is the question. I am unable to understand how the last step of the answer is done
Can someone help me
2 hours later…
8:03 AM
@Amr I am an engineer
8:22 AM
Can someone help with my above question
What the difference between the ramp and saw tooth function , like how to write them ?
Should we include the frequency parameter also
IN the function
9:20 AM
There is an interpretation of $G$-structures as equivalence classes of the metric tensor, which apparently relates to their interpretation as Cartan geometries via their stabilizer groups being subgroups, but how does one compare two such $G$-structure when they don't even have the same model?
You can relate the metric structure and conformal structure by the equivalence $f^*g = e^{2u}g$, but how does this work out on a bundle level?
9:32 AM
Most sources I can find on the topic assume the model space to be $\mathbb{R}^n$ with the stabilizer being a subset of $\mathrm{GL}(n)$, but that doesn't seem to be the case for the conformal or projective group
I guess I can maybe consider the projective space of the $\mathbb{R}^n$ in question for the projective structure, but I'm not quite sure how it works out for the conformal structure with the Möbius space
1 hour later…
10:44 AM
Is there a standard name for the following graph theory problem? --> Given a directed graph, find the vertex (vertices) from which the largest number of other vertices are reachable.
One might call it "largest out-component"
11:18 AM
@Szabolcs so largest degree?
3 hours later…
1:56 PM
@LeakyNun No. "Reachable" does not mean "connected by an edge". t is reachable from s if there is a path from s to t.
Largest degree vertex in the transitive closure.
2 hours later…
3:48 PM
@Szabolcs then you shouldn't think about vertices right
take all the connected components, and find the one with the most vertices
then every one of those vertices will be a vertex you want
the graph is directed @Leaky
I see
But still there should be an O(n) algorithm adapting a connected components algorithm I would guess
4:13 PM
you probably can reduce the graph first (by merging nodes that are part of a cycle), then you get a tree and are just asking for its depth
well, a forest
if you're not limited in memory, you prob can do this in O(n)
4:33 PM
oh wow, I love the orthogonal group!
Can someone ELI5 ordinals? I don't understand the Wiki.
@UnderMathUate There's two ways to think about ordinals
The first is the idea that, after every set of ordinals, there's one more
So it starts with 0, 1, 2, 3, 4, …
but then, from that property, there's another ordinal after the set {0,1,2,3,4,…}. Call it ω (omega)
and then there's another ordinal after {ω}, called ω+1
So the ordinals go 0, 1, 2, 3, …, ω, ω+1, ω+2, …
but then there's another ordinal after all of those. That's ω+ω, or ω·2
(for annoying reasons we don't call it 2ω)
Then ω·2+1, ω·2+2, …, ω·3, ω·3+1, ω·3+2, …, ω·4, …, …, ω^2
You can see a poster here:
4:50 PM
This is a pretty poster ;-;
It makes sense now. Ordinals are ways to order sets and there are infinitely many ways to do that.
By the way, I said "After every set of ordinals there's another ordinal"
"What about the set of all ordinals?"
This is called the Burali–Forti paradox, and it's resolved in a similar way to the Russell paradox ("The set of all sets that do not contain themselves") - you use a system where certain sets are forbidden (they're "too big to be sets")
Such a system is called "a set theory", and one popular one is called "ZFC set theory"
Burali-Forti is one person with a hyphenated last name, not two people, by the way
Oh wow
There's another way to define ordinals, by the way, as "order types of well-ordered sets".

A set is well-ordered if every subset has a minimum. Two sets have the same order type if there's an order-preserving bijection between them.
The ordinal ω+1, for example, is the order type of
{0, 1/2, 2/3, 3/4, … 1}.
It's also the order type of
{0, 1, 2, 3, … ω}.
Each ordinal is the order type of the set of smaller ordinals.
4:54 PM
Do you think Halmos' Naive Set Theory is accessible to an undergrad?
I've never read it
I mean, my recommendation anyway is "Get the book and see how far in you get"
(unless it's very expensive - I don't know how much it is)
It isn't expensive at all. I can get it for less than $10 on Amazon.
I think trying to read books you're "not ready for yet" is a good idea because, (a) sometimes not all of the "prerequisites" are actually required, and (b) if there's something you don't get, or something that requires knowing some background, now you have motivation for learning that background
(and a lot of math is very "bring your own motivation" so take it where you can find it)
Ok, appreciate the advice. I'll buy it then.
@UnderMathUate The person who made that poster, Joel David Hamkins, also has some books out that you may appreciate
5:04 PM
@AkivaWeinberger yes, very much so
I assume you meant to tag @UnderMathUate
wrong ping
yeah, misclicked
it's written for undergrads. i never 'learned from' it, but having taken a course in set theory, it seemed good when i did look at it
5:05 PM
I've read Halmos' Naive Set Theory and wholeheartedly recommend it
except for the proof of Zorn's Lemma in there, supplant that with one from elsewhere
i learned from enderton's "elements of set theory," which i thought good, but i think it's less accessible for self-study than halmos
it helps to have a prof on that one
i also enjoyed supplementing with halmos
it's good to learn transfinite induction at some point in your life
it's a very fun thing to do, especially when it's inappropriate
5:50 PM
By the way, @UnderMathUate
The most striking example of ordinals being used outside of set theory is in the theory of braids.

Braids have something called the Dehornoy (pronounced "de orn-WAH") ordering.
If you restrict to braids where all the overcrossings are the same direction, those braids turn out to be well-ordered under the Dehornoy ordering (recall that "well-ordered" means every subset has a least element).
That set of braids turns out to have order type
where n is the number of strings.
> the most striking example I KNOW OF of braids being used outside of set theory
^is what I meant to write
A note on the name. You may have also heard of "cardinals". "Cardinals" and "ordinals" are similar concepts with similar names, so it may be hard to keep them apart. The important thing is that ordinals refer to order, hence the name
(Another use of ordinals is in something called the "fast-growing hierarchy", used to define functions that get large very quickly)
chat.stackexchange.com/transcript/message/59804886#59804886 this is not quite right, but you get a graph in which it's easy to compute anyway
6:33 PM
Is it possible for mutually orthogonal vectors to not intersect? And if they do, what is the proper term?
mm, what do you mean by 'intersect'?
I mean as the case in coordinate system which they do intersect but should this be always the case?
typically vectors are drawn coming out from origin so in this sense they all intersect
so you need to clarify what you mean by intersect in another sense
Let's say two lines (one vertical and one horizontal) in 3D and assume there is a distance between them. Can we describe them as orthogonal vectors?
Vectors are interpreted as not having a base point. They're simply a direction and magnitude.
If I draw an arrow over here on the left, and I draw another arrow over there on the right with the same direction and size, they are considered to be equal vectors
Vectors are "free to slide around"
6:40 PM
But what is the proper term if they form coordinate system?
the definition of orthogonality in vectors is their dot product is zero
if you're speaking of line equations, orthogonality doesn't have the same definition
From wiki, Cartesian coordinate is defined by an ordered pair of perpendicular lines. What is the proper term other than saying two perpendicular lines?
@Slereah that is my question. Orthogonality means the angle between two vectors is 90 but this definition doesn't say anything about whether they are intersect or not.
you have to distinguish line equations from vectors
a vector in a cartesian plane is perfectly well defined by an ordered pair, but this isn't enough for a line equation
6:48 PM
@CroCo Linearly independent?
@Slereah linearly independent is not necessarily perpendicular.
you have to distinguish vectors from graphs of line equations
1 is a vector, the ordered pair (2,4) is a vector, the ordered tuple (3,6,10,2,2) is a vector, but none of those are line equations or their graphs
@shintuku then how can I describe three orthogonal unit vectors that where their start point is the same.
vectors have no notion of "starting point", no relation to that concept
you can describe their orthogonality by saying their dot product is 0
why does the reisz representation theorem for sequence spaces imply that that the dual of $l_1$ is not separable? I know that $l_{\infty}$ is not seperable but i'm asking as to why RRT implies that
6:57 PM
@CroCo you can however define segments of line equations by limiting their domain. for example, in the cartesian plane, the graphs of the line segments [0,a] and [b,0], for 0<a,b<1
Those don't look like lines to me at all.
line segments heh
assume these vectors as unit vectors.
what is the proper term, if any, the describe this?
No, that makes no sense.
Vectors should have their tails at the origin.
You're still confusing language badly.
@CroCo three dimensional cartesian plane with line equations going through origin, and for some reason there's a little arrow at the tip, probably in order to deliberately confuse the students hehe
7:00 PM
@shintuku but this definition for 3D.
I'm asking in general
Well, the arrows indicate the direction of increasing values along the axes — otherwise called orientation. Nothing to do with vectors.
oh right ted
@CroCo that seems like a three dimensional cartesian coordinate system with increasing values in the direction pointed
maths, the term 'riesz representation theorem' is so overused that i'm not quite sure what you mean. is it just the identification of the dual of ell^1 with ell^infty? people do sometime use "RRT" for that (or more generally for some 'concrete' identification of the dual of a banach space)
I'm looking for a proper term to define what is a frame which is used in robotics to refer to three unit vectors where their tails meet at a specific point
If you don't want that point to be the origin — in which case we call it a basis for $\Bbb R^3$ — you can talk about a frame at a point $P$.
7:05 PM
@TedShifrin is it always a basis orthogonal ?
No, you have to say orthogonal or orthonormal basis or frame.
so a frame is an orthonormal basis ? right?
No, you have to specify an orthonormal frame
Frame is just another word for basis.
Often used when it is a moving frame and changing as you move from point to point.
@TedShifrin thank you so much. This is what I'm looking for exactly.
8:10 PM
@AkivaWeinberger I'll have to look in them. I haven't heard of these concepts before. Well -- I have seen chains. I guess that concept may be similar to the braids.
8:54 PM
@leslietownes yes, it's just the identification
maths: OK. so the dual of ell^1 is not separable because (by that result) it is isometrically isomorphic to ell^infty, which is not separable (i assume this is known?).
and separability is invariant under isometric isomorphism.
it's probably enough to know that the dual of ell^1 contains an isometrically embedded copy of ell^infty; you don't need the full identification of the dual.
but you do need to know or otherwise prove that ell^infty isn't separable.
Dropping some ell's
hold this ell
under the tongue
If I had some ell this cleaning job would go a lot quicker. I have to do like 20 deep-cleaning tasks around house b/c we have 12 cats now and my grandma is moving in. I've already been constructing a cat kennel with father for the week. The kitchen alone is going to take me another 1.5 hours
9:20 PM
that is a lot of cats
@TedShifrin. @leslietownes. @robjohn. @PM2Ring. Thank you all for your help during my math exam. I passed it. It was very hard without you guys.
9:36 PM
@Avra That's pretty stonks. You done for the semester?
10:30 PM
They're done talking, apparently. :)
Greetings to the toy of the munchkin.
11:19 PM
hi ted
ah, the toy speaks
speak and steak do not rhyme
bough, cough, dough
English pronunciation is absurd.
11:35 PM
When you cough, you must dough your hat. I don't want to discuss boughing, however, as that is a private matter.
You use self-rising flour when you dough your hat?
Certainly. And then I tie the dough into a bough, like a pretzel.
Bows and boughs have different topology.
And coughs say moo!
I say neigh to that.
11:48 PM
i'm through with this
if copper were here he'd say something rough
Oops, I forgot to represent the rough side.
coughs say mough

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