Logic - 2018-03-09

user21820

05:06

@WhatsThePoint Correct. So the answer is (((D∧¬O)∧A)∨(M∧¬(O∨¬A)))→S, which is equivalent to famesyasd's (((D∧¬O)∧A)∨(M∧(¬O∧A)))→S. Note that we often use the following precedence rule to reduce the number of brackets. Bracketed expressions are still evaluated first, and the boolean operations are processed in the following order: ¬,∧,∨,→. You are in fact already using the precedence of ¬ over the rest. Using this precedence rule, the answer can be written as D∧¬O∧A∨M∧¬(O∨¬A)→S.

For your schoolwork, you may not be permitted to use this precedence rule, but I hope it is clear why conventionally people adopt it.

Specifically, it is usually the case that we want to express a conditional statement where the condition and the consequence are compound statements involving cases or criteria/results. Cases involve "or", while criteria/results involve "and", and so that is why we like "implies" to have least precedence. And usually we have multiple criteria/results within each case, and so we like "and" to have higher precedence over "or".

Your example about the alarm illustrates this phenomenon very well, even though it is extremely convoluted. In practice people will instead say "The alarm in the server room will sound and send an SMS if the alarm is activated and the override button is not pushed and either the security door is opened or motion is detected." which translates directly to "A∧¬O∧(D∨M)→S".

WhatsThePoint

08:10

ah thank you, I think I am starting to understand it now, the topic goes a bit over my head really if I'm being honest

user21820

10:11

@WhatsThePoint It's okay. Just ask about anything that is unclear to you or that you want to know. Aim for 100% understanding, and you'll get there!

famesyasd

10:41

@user21820 is $ f(t)=(sin^3t,cos^3t), t\in R$ (astroid) a function? it does not pass the vertical test

it passes main function criterions though

ah no, it's not

user21820

@famesyasd Technically, what you wrote is not even a mathematical object, so it cannot be a function or anything. I want to be technical because everyone who doesn't stick to technical precision has similar kinds of misconceptions.

Let me state the precise facts:

(1) There exists a function f : R → R^2 such that ( f(t) = (sin(t)^3,cos(t)^3) for every t in R ).

famesyasd

so that's what we mean when we say, that something is a function or that something is not a function?

user21820

(2) For every function f : R → R, the graph of f is the set { (x,f(x)) : x in R }, and obviously you can see that the graph of f has a unique pair (x,y) for each x in R.

@famesyasd Wait.

(3) A function is not at all the same as its graph.

(4) A function is simply an object f with a domain S that for each input x in S specifies the output of f on x.

You can treat (4) as the definition of a function. It should then be obvious why (1) is true, because I literally have specified the domain and the output on each input in the doman.

@famesyasd: So based on (1) to (4), I will ask you some questions and you answer me. Let f : R → R^2 such that ( f(t) = (sin(t)^3,cos(t)^3) for every t in R ).

Is f a function?

Is f(t) a function?

Is f(u) a function?

Is f(0) a function?

Is there a set D and function g : D → R such that { (x,g(x)) : x in D } is the same set as { f(t) : t in R }?

famesyasd

12:25

okay, I'm still writing :D I'll post after I eat

All I can say is that I started understanding something but I can't think further I need to eat

famesyasd

15:16

@user21820 are you here?

user21820

@famesyasd I will be back later. Take your time!

famesyasd

@user21820 I think I got it :D

user21820

Great! =)

Tell me all your answers and I will check them when I return.

famesyasd

@user21820 kay

user21820

16:14

@famesyasd: Okay I'm back for a while. Just to let you know, I updated my post about introductory texts. I guess it's too risky for me to link to references that are not written by professional logicians haha, so I removed the link to forallx. Oh well. Anyway, Noah just posted a new answer recommending Boolos' logic textbook.

famesyasd

@Wait! actually I have checked those rules and they are ok! He protects from that 1<0 kinda thingy by disallowing for $\forall$ introduction rule to be introduced with assumptions (those that are separated by underlines) and axioms act exactly like that! so I think he's actually got it quite nice with those rules

user21820

@famesyasd No you're wrong about that. Axioms are not assumptions.

Also, as I stated earlier he used the term "constant" in inconsistent ways.

famesyasd

16:32

@I don't see why not, aren't they like initial assumptions?

user21820

No they are not. A formal system may have infinitely many axioms. An assumption in a proof is not an axiom but a temporary assumption, under which you can deduce statements depending on that assumption. This is the whole basis for Fitch-style in the first place.

famesyasd

@hmm, but even if there are infinitely many axioms we can use only finite of them everytime so we can think of them as having only finite amount of axioms each time, and for the temporary assumption, but we can think of axioms also as temporary assumptions except that they have much much longer lifespan:) and then we look at what we can deduce from them just as with initial temporary assumptions, I know that I might be twisting here but I honestly can't see why this twisting is bad

user21820

You don't have to believe me if you don't want to, but I can easily prove you wrong by a challenge: Go and write a program that will check any Fitch-style proof and output "valid" if it is valid and "invalid" otherwise. If you succeed, you will totally agree with what I said about forallx. If you do not write this program, then trust my judgement.

And you are not the only one who has been confused by forallx. 2 other users that I taught in this chat-room were confused by something else, where forallx was inaccurate in its definitions. So I switched completely to the Fitch-style system I had linked you to, and taught them from scratch, and they got a much clearer understanding.

You can see our discussions here:

Fitch-style propositional logic & Examples

Dec 19 '17 at 12:56, 4 minutes total – 7 messages, 1 user, 0 stars

Bookmarked Jan 2 at 14:12 by user21820

Fitch-style first-order logic & Examples

Jan 5 at 14:05, 1 hour 22 minutes total – 143 messages, 3 users, 0 stars

Bookmarked Jan 27 at 3:12 by user21820

Sketch of semantic-completeness theorem for propositional logic

Dec 18 '17 at 13:27, 1 hour 51 minutes total – 154 messages, 2 users, 0 stars

Bookmarked Dec 19 '17 at 13:07 by user21820

The rules are only slightly different from the ones in the linked post, partly to satisfy their philosophical objection to having an inbuilt sentence "false".

famesyasd

mmm okay, well then, I believe you for now:) Don't have time to check and try this out myself

user21820

In lieu of actually doing that, you can simply try to use the system to prove things in PA.

Do you want exercises to try?

I will check your answers if you do them.

Then I can point out where you're using Fitch-style wrongly. You don't have to specifically use my system or his system, but it must make sense to me in terms of having contexts and every sentence true in its context.

The only thing is that you must be 100% consistent.

famesyasd

16:49

Not now, I don't have time this week (or maybe a couple) xd I have to pace through some of algebra/differential geometry and analysis now :D Later! Definitely!

user21820

@famesyasd Ah okay! Just let me know when you want them.

I'll be off for now.

See you!

famesyasd

wait!

"Is f a function?" Yes, I don't know how to prove that technically nicely here but it is clearly a function now that I have pictured it

Now I see that I confused the actual plot of the function f (which I actually have no idea how to visualise. Well, I constructed two ways, one of them being with some basis in R^3, one of the lines would be for t (arguments) and the other two would be for sin^3(t) and cos^3(t) respectively so we can identify those points in R^3 (t, sin^3(t), cos^3(t)) with (t, (sin^3(t), cos^3(t)) Well, basically, that would look like cylindrical astroida or something like…

xd

user21820

@famesyasd Good! For reference, the answers to my questions are: f is a function. f(t),f(u) are meaningless because t,u are undefined! f(0) = (0,1) which is just a pair in R^2 and not a function. And no to the last question, and here is how to prove it rigorously:

famesyasd

what? f is not a function? why?

user21820

Accidental mental slip sorry; I edited my message.

famesyasd

16:58

phew

xd

user21820

Given any set D and function g : D → R:
  Let S = { f(t) : t in R }.
  Let T = { (x,g(x)) : x in D }.
  f(π) = (0,1) and f(−π) = (0,−1).
  Thus (0,1),(0,−1) in S.
  If S = T:
    (0,1),(0,−1) in T.
    exists x in D such that (0,1)=(x,g(x)).
    exists x in D such that (0,−1)=(x,g(x)).
    Let p in D such that (0,1)=(p,g(p)).
    Let q in D such that (0,−1)=(q,g(q)).
    p = 0 = q.
    1 = g(p) = g(q) = −1.
    Contradiction.
  Therefore S ≠ T.

famesyasd

Have you proved that with the rules from your post?

user21820

It is more or less compatible, once you add set builder notation and axioms for them.

The point is that it's the logical structure that is the most important, and the other details don't matter too much. Here the logical structure is basically a proof by contradiction within the context given any set D and function g : D → R.

And also the meaning of set membership is clear from the above proof; it translates into an existential statement. Also, note that we must use fresh variables p,q; after we use p for one exists-elim we cannot use p again for the next one. This should make obvious sense to you; if you use p to refer to some witness of a true existential sentence you cannot expect it to refer to a witness of another true existential sentence.

It turns out that in this particular proof we do prove that p = q, but the point is that it needs proof, not hand-waving.

Anyway I'm going off. Leave a message if there is any point you don't fully understand. See you!

famesyasd

17:13

@user21820 okay, no actually your proof follows what I had in mind pretty 100% lol, the point on new variables I'll grasp later, what I can ask also, is that if you compute derivatives of f at those 4 curly points (at those t's where f(t) are those curly points) then it's vanishing there! what does it mean? I don't have any intuition behind derivatives and it's values in higher oorder dimensions but from 1 variable calculus it should not have existed at the vertucal ones ate least I think

it should have been something like $\infty$ there

Leaky Nun

22:29

12 hours ago, by user21820

(3) A function is not at all the same as its graph.

@user21820 actually they are :P

Fitch-style propositional logic & Examples

Fitch-style first-order logic & Examples

Sketch of semantic-completeness theorem for propositional logic

Transcript for

♦ $Mathematics$ Logic

Fitch-style propositional logic & Examples

Fitch-style first-order logic & Examples

Sketch of semantic-completeness theorem for propositional logic

Transcript for

♦ Logic

♦ $Mathematics$ Logic