Ok, here's my logic for the day.

∵ Twilight is a Book.

∵ Books are Dead Trees.

∵ Fluttershy is a Tree

∴ Twilight is Dead Fluttershy

There is logic in what he says

EDIT: Wait - one must first prove that Twilight is a book

Ok, proving Twilight is a book.

∵ Twilight Reads a lot of books

∵ What you do defines you

∵ Fluttershy is good with nature and is a tree

∴ Twilight Sparkle is good at reading and is a book Q.E.D.

More comprehensive version coming when I have too much free time on my hands, and decide to work on it.

More comprehensive? I'll help.

Start with this:

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.


You can now use addition in your proof.

Now, I understand that Peano's axioms are indeed a good starting point, but I have to add a remark concerning your conclusion.

What you define as 'addition' isn't the operation usualy called by that name. It's only a function f:N²->N . Since you define 2, it doesn't have 'numerical' value as well, because it could be any mathematical object as well.

If you want to have a classic addition, you might want to prove that this operation is:
- commutative (ie a+b = b+a)
- associative (ie a + (b+c) = (a+b) + c)
- possession of an additive identity (ie x where a+x=a)

That's the third point that bugs me, right ? Because you took the Peano's version where 0 is not in the set, so you can't prove that. It means everything you said is correct, but define an operation where (N,+) is not a group, but rather "a set and an operation".

It totally works with many problems, but just pointing it out, so rarity can use it the way it should.

/completlymissingthepoint

While I probably could adjust this to address Luna's concerns,

1. That's a lot of work and I'm busy/lazy
2. Making Rarity work having only partially proven addition should be more amusing

Also, in the way of "assuming nothing," I have remembered a great quote from textbook: "So far, we have assumed reality. Henceforth, we will no longer constrain ourselves."

Unfortunately, it was just talking about real numbers

Rarity wrote:

Ok, here's my logic for the day.

∵ Twilight is a Book.

∵ Books are Dead Trees.

∵ Fluttershy is a Tree

∴ Twilight is Dead Fluttershy

The best thing is; this is logically correct. xD (I have just recently learned a bit on formal logic in my philosophy classes xD )

The Jack wrote:

Rarity wrote:

Ok, here's my logic for the day.

∵ Twilight is a Book.

∵ Books are Dead Trees.

∵ Fluttershy is a Tree

∴ Twilight is Dead Fluttershy

The best thing is; this is logically correct. xD (I have just recently learned a bit on formal logic in my philosophy classes xD )

Yes, it is logically correct. Problem is that as I have not proven Twilight is a Book (Or Fluttershy as a Tree for that matter, or even that Books are Dead Trees. Man, I've got a lot of work still ahead of me) It's flimsy logic at best. Don't worry, I'm trying to get it working to a decent enough level that I can prove that Twilight is a Book, Fluttershy is a Tree, and Books are Dead Trees. Hopefully in a sufficiently complicated matter that makes anyone who looks at it wonder why I went to all that work.

It's also not logically correct.
It's stating (A ∈ C) ∧ (B ∈ C) ∴ A = B with
A = Books
B = Fluttershy
C = Trees or tree products

This can be easily disproven:
A = 1
B = 2
C = ℝ

Rarity wrote:

The Jack wrote:

Rarity wrote:

Ok, here's my logic for the day.

∵ Twilight is a Book.

∵ Books are Dead Trees.

∵ Fluttershy is a Tree

∴ Twilight is Dead Fluttershy

The best thing is; this is logically correct. xD (I have just recently learned a bit on formal logic in my philosophy classes xD )

Yes, it is logically correct. Problem is that as I have not proven Twilight is a Book (Or Fluttershy as a Tree for that matter, or even that Books are Dead Trees. Man, I've got a lot of work still ahead of me) It's flimsy logic at best. Don't worry, I'm trying to get it working to a decent enough level that I can prove that Twilight is a Book, Fluttershy is a Tree, and Books are Dead Trees. Hopefully in a sufficiently complicated matter that makes anyone who looks at it wonder why I went to all that work.

Nonono; it's not flimsy logic. Infact, it's a perfectly viably logical syllogism, if I remember the name correctly.
The problem is, it's not neccesarely TRUE. xD
For that, you'll have to prove that Twilight is a book and that Fluttershy is a tree. If those two are correct, then the conclusion MUST be correct. =D
(I never knew I'd actually USE this stuff they teach us. I mean, WTH? Why do they have a philosophy course that's mandatory in a physics education? xD )

Instead of proving that there is only one element in Tr, we could also prove that T and F are the same element in Tr, that is that Twilight and Fluttershy are the same object.

Okay, really now.

You guys are making me feel like I have learned nothing in my years of public education.

Good new, that's college level discrete mathematics, so you shouldn't feel bad.

Well, you shouldn't feel bad about this thread at least. I don't know what else you've been up to.

Rydel wrote:

Good new, that's college level discrete mathematics, so you shouldn't feel bad.

Well, you shouldn't feel bad about this thread at least. I don't know what else you've been up to.

Umm I'm up to pre-calculus, aka Math Analysis, aka Anal Math. Almost there! Right?