Endless Paradigm - [split] Anime VS Manga

Assassinator Wrote:
Goshi Wrote:The game's [ero's] main purpose was to improve the sales. The translation patch has a non-ero version anyway but the text still hints towards them anyway. Kohaku's storyline in particular. There was a work-safe port of FSN though but correct me if I'm wrong.
Improve sales... But really, if people want to see the ero that much, they should just go download some hershey kisses or something, buying a visual novel is probably the most roundabout and expensive way to do it.

It worked with Kanon though. As for Kohaku's storyline well... uh... hmm..... Kohaku's storyline.

Quote:Seems like u made an account in Beast's Lair. One day i was bored and looking thru some random threads there, saw that familiar sig of yours, then looked at the username.

Yeah I did. Don't go there too much though.

Goshi Wrote:It worked with Kanon though. As for Kohaku's storyline well... uh... hmm..... Kohaku's storyline.

Havn't actually seen Kohaku's storyline. is it really that bad?

Goshi Wrote:Yeah I did. Don't go there too much though.

Neither do i. I don't even have an account.

I just go there to get manga and visual novel translations mainly. Sometimes i read a bit of random threads that seem interesting, but overall i spend little time there.

Sparker Wrote:
Assassinator Wrote:
ZiNgA BuRgA Wrote:Obviously I have heard of it. I was just being a Goshi.

Watched it? or just heard.
You need to read a little bit closer :p

:@

He said he "heard" it, but that does not exclude the fact that he could have watched it too.

Watched and Heard are not mutually exclusive, so by saying that he heard, it tells nothing about whether he watched or not.

Put it mathematically, let A = he heard. B = He watched.

Wee are given:   1.) A is true
                      2.) B -» A.

U are stating:    A -» ~B.

If u can somehow mathematically prove how that would work, then I'll willingly donate all my assets to you, and be your slave for the rest of my life.

Assassinator Wrote:
Goshi Wrote:It worked with Kanon though. As for Kohaku's storyline well... uh... hmm..... Kohaku's storyline.

Havn't actually seen Kohaku's storyline. is it really that bad?

Goshi Wrote:Yeah I did. Don't go there too much though.

Neither do i. I don't even have an account.

I just go there to get manga and visual novel translations mainly. Sometimes i read a bit of random threads that seem interesting, but overall i spend little time there.

Sparker Wrote:
Assassinator Wrote:
ZiNgA BuRgA Wrote:Obviously I have heard of it.  I was just being a Goshi.

Watched it? or just heard.
You need to read a little bit closer :p

:@

He said he "heard" it, but that does not exclude the fact that he could have watched it too.

Watched and Heard are not mutually exclusive, so by saying that he heard, it tells nothing about whether he watched or not.

Put it mathematically, let A = he heard. B = He watched.

Wee are given:   1.) A is true
                      2.) B -» A.

U are stating:    A -» ~B.

If u can somehow mathematically prove how that would work, then I'll willingly donate all my assets to you, and be your slave for the rest of my life.

1. A is false

therefore

A -» ~B

because false implies anything is true.

:@

and wee all know Zinga is a liar.

_VEndeTta Wrote:1. A is false

therefore

A -» ~B

because false implies anything is true.

:@

and wee all know Zinga is a liar.

Ok.

1. Wee are assuming A is true, as per the question, thank you.

2. Even if A is false, It makes no difference at all. The logical argument still fails.

Regardless of whether A is true or false, u still got to prove

B -» A.

|
V

A -» ~B.

It still doesn't work.

Your question is stupid. There are many cases in which (B » A) » (A » ~B) is true as can be seen here:

A | B |  B » A  |  A » ~B | (B » A) » (A » ~B)
T   T        T             F                    F
T   F        T             T                    T
F   T        F             T                    T
F   F        T             T                    T

Is it a tautology? No. But it only fails when both A and B are true.

There. It has been proven by cases. Kthxbai.

Since A is true, your case only works if B is false, which pre-supposes that I haven't watched...

But since I'm a liar, B must be true, thus A and B is true, thus the above does not work.

_VEndeTta Wrote:Your question is stupid. There are many cases in which (B » A) » (A » ~B) is true as can be seen here:

A | B |  B » A  |  A » ~B | (B » A) » (A » ~B)
T   T        T             F                    F
T   F        T             T                    T
F   T        F             T                    T
F   F        T             T                    T

Is it a tautology? No. But it only fails when both A and B are true.

There. It has been proven by cases. Kthxbai.

If it is not a tautology, then it fails. It has to be true for all cases for the logical argument to be correct. U can't just say, it's true for some values, so it's correct, and ignore the other values.

Lets take your reasoning and put it in an example.

Statement:   x is always equals to 10.

Proof:
There is a case in which x is equal to 10.

           |    x     |  Is x equal to 10?
Case 1     <10              No
Case 2     =10              Yes
Case 3     >10              No

Is it a tautology? No. But it only fails when x<10 and when x>10.

There. It has been proven by cases. Kthxbai.

Does that mean i proved that "x is always equal to 10?" NO.

You have to prove it correct for all cases in order to prove a mathematical statement to be correct.

_VEndeTta Wrote:Your question is stupid.

I didn't intend this for people to try to prove it, because it's incorrect, thus unprovable.

I intended it to just say to Sparker that just because Zinga wrote that he "heard" it, it does not exclude the possibility that he could also have seen it.

Thus justifying my question to Zinga "Watched it? or just heard it?"

Just to clear everything up:

Ok, I realize u intend your original response (the one about Zinga lying) as a bit of humor, and i took it seriously.

And I'll apologize for that.

(and I'll remove that "thou hast failed" picture, which was intended as nothing more than a joke)

And instead, I'll put it here...
[Image: 663n61z.jpg]

Joking man. I apologise again.

Code:

			#define x 10
#include <someshit>

int main()
{
 if(x==10)
  printf("x is always 10!!!");
 
 return death;
}

The "mathematical" formula Assassinator made sounds like something the void would spit out.

Assassinator Wrote:If it is not a tautology, then it fails. It has to be true for all cases for the logical argument to be correct. U can't just say, it's true for some values, so it's correct, and ignore the other values.

Lets take your reasoning and put it in an example.

Statement:   x is always equals to 10.

Proof:
There is a case in which x is equal to 10.

           |    x     |  Is x equal to 10?
Case 1     <10              No
Case 2     =10              Yes
Case 3     >10              No

Is it a tautology? No. But it only fails when x<10 and when x>10.

There. It has been proven by cases. Kthxbai.

Does that mean i proved that "x is always equal to 10?"       NO.

You have to prove it correct for all cases in order to prove a mathematical statement to be correct.

Lol. I was in a really bad mood when I responded to your last statment, normally I just ignore it when people don't
get my amazing jokes. ;P

But, I took your question to be, does (B » A) imply (A » ~B). Which is a rediculus statement, its like asking does
X + Y = 50 given X = 34, its just not a question you ask, instead you ask what must y be such that X + Y = 50 given X =34
I think what you wanted people to prove was.

A
B » A
--------
A » ~B

Which is obviously wrong. As u can see by the truth table. Harr. I liek the mathorz.