Re: Ritz ballistic emission theory material

Hexenmeister a écrit :
"YBM" <ybmess@xxxxxxxx> wrote in message news:4440fca5$0$11338$636a55ce@xxxxxxxxxxxxxxx
| Hexenmeister wrote :
| > Hey moron, xor is a binary operator, you stupid ***, just like +, -, * and /
| > are binary operators. They don't "imply" at all.
| | so what ? did I say otherwise ?

Yes, you did, you dumb ***.

Where ? How ? Quote ?
I wonder if you only understand the meaning of words like "binary
operator" or "imply"...


| btw => and <=> is as well a binary operator
| | > | Even worse, are you pretending that (False xor False) is True ??
| > | > Hey shithead, NOT FALSE = TRUE in Boolean Algebra, so | > (NOT FALSE) AND (NOT FALSE) = TRUE, haemorrhoid. | | are you saying that A xor B is ((not A) and (not B)) ??

No, I'm pretending it. What I'm saying is you are fucking ineducable idiot!!
*** off.

you're loosing you cold blood. You know there is nothing bad in
recognizing a mistake, especially in science.

I've been a bit optimistic by trying to convince you by undoubtful
references and checkable propositional derivations. Let's take another
path to make you learn something, so that you didn't lose completely
the few time you'll had spent here before your death.

One : in order to refute a propostion of the form P => Q as well as
the assertion that P implies Q, one only has to find a case where P is
True and Q is False.
(agreed ?)

Two : for propostions of the form (A xor B) or (A or B) there are
only 4 different cases : (A is true, B is true) (A is true, B is false)
(A is false, B is true) (A is false, B is false)
(agreed ?)

Three : By looking at these four cases, if one found one where (A xor B)
is True and (A or B) is False, one could then say - AS YOU DID - that
it is not true that (A xor B) implies (A or B), otherwise if no such
case exists, one should then admit that (A xor B) implies (A or B)
(agreed ?)

Please don't forget to read, wait a bit and *think* before fooling
yourself.
.

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