Commercial Signatures

[trythil]

A mathematical proof?

If ANIME may be placed on the GD,
but ANIMATION may not be placed on the GD,
then ANIME ≠ ANIMATION?

No, just that the two aren't equivalent. Anime is a subset of animation, but animation is not a subset of anime. That is to say, whilst all anime is animation, not all animation is anime. So if the Golden Doughnut required the footage to be anime, there would be no inconsistency to ban other animation.

Damn, I knew that there had to be a use for all that set theory I studied

[Kenmakiryu]

Well, I meant no offense by calling it the 'doughnut deal'.

But, I'll probably keep calling it that since I always put 'deal' at the end of anything that seems like a deal. Anything good, I always add 'deal' to.

[Ruinku]

alright call of the sniper he's on the good side :p

[EarthCurrent]

No, just that the two aren't equivalent. Anime is a subset of animation, but animation is not a subset of anime. That is to say, whilst all anime is animation, not all animation is anime. So if the Golden Doughnut required the footage to be anime, there would be no inconsistency to ban other animation.

*untrusting* Let's see the formula...

[trythil]

No, just that the two aren't equivalent. Anime is a subset of animation, but animation is not a subset of anime. That is to say, whilst all anime is animation, not all animation is anime. So if the Golden Doughnut required the footage to be anime, there would be no inconsistency to ban other animation.

*untrusting* Let's see the formula...

Anime is a subset of animation, but animation is not a subset of anime.

A and B are equivalent if and only if A is a subset of B and B is a subset of A.

At least that's the definition I learned in discrete math. Linguists seem to define sets A, B equivalent iff they contain the same number of elements, which isn't the same thing.

[Corran]

Dokool and Krypton-Knight seem easy to tick off with trivia name calling.

[Deaths_ally]

A and B are equivalent if and only if A is a subset of B and B is a subset of A.

At least that's the definition I learned in discrete math. Linguists seem to define sets A, B equivalent iff they contain the same number of elements, which isn't the same thing.

*head spins.. passes out*

[trythil]

A and B are equivalent if and only if A is a subset of B and B is a subset of A.

At least that's the definition I learned in discrete math. Linguists seem to define sets A, B equivalent iff they contain the same number of elements, which isn't the same thing.

*head spins.. passes out*

It's pretty simple, actually.

Take sets A, B from the set of all positive integers:

A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}

Is A a subset of B? Yes, A is a subset of B. (It's not a proper subset of B, since A is B, but a subset can include the entire set.)
Is B a subset of A? Again, yes. So A = B, and B = A.

Let's try this again:

A = {1, 2, 3}
B = {1, 2, 3, 4, 5}

Is A a subset of B? Yes, because A is {1, 2, 3}, and B is {1, 2, 3, 4, 5}.
Is B a subset of A? No, because B contains elements that are not in A -- namely, 4 and 5. So A is not equal to B, and B is not equal to A.

This isn't really a proof of the statement "sets A and B are equal to each other if A is a subset of B and B is a subset of A", but it does demonstrate how the statement works.

[Ruinku]

what that means in english
animation is a catagory and anime is a specific type of animation

basically its like two cups
animation is a big cup and anime is a smaller cup
you can fit the small cup into the big cup but you can't fit the big cup into the smaller cup

[Mr Pilkington]

A and B are equivalent if and only if A is a subset of B and B is a subset of A.

At least that's the definition I learned in discrete math. Linguists seem to define sets A, B equivalent iff they contain the same number of elements, which isn't the same thing.

*head spins.. passes out*

It's pretty simple, actually.

Take sets A, B from the set of all positive integers:

A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}

Is A a subset of B? Yes, A is a subset of B. (It's not a proper subset of B, since A is B, but a subset can include the entire set.)
Is B a subset of A? Again, yes. So A = B, and B = A.

Let's try this again:

A = {1, 2, 3}
B = {1, 2, 3, 4, 5}

Is A a subset of B? Yes, because A is {1, 2, 3}, and B is {1, 2, 3, 4, 5}.
Is B a subset of A? No, because B contains elements that are not in A -- namely, 4 and 5. So A is not equal to B, and B is not equal to A.

This isn't really a proof of the statement "sets A and B are equal to each other if A is a subset of B and B is a subset of A", but it does demonstrate how the statement works.

This is also useful in proving that 2 wrongs do indeed make a right.