Proving that nothing does not exist

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The discussion presents a mathematical argument aimed at proving that "nothing" does not exist, using logical propositions and definitions. It outlines a series of hypotheses that lead to the conclusion that the absence of everything implies the existence of everything, ultimately resulting in the assertion that "nothing" cannot be true. The argument employs the laws of excluded middle and non-contradiction to support its claims. Some participants question the validity of the demonstration, suggesting it may be a play on words rather than a serious proof. The conclusion drawn is that the space cannot be devoid of existence, reinforcing the idea that nothing does not exist.
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I applied some mathematical view to the daily language while studying demonstration.

Proving that nothing does not exist

Consider the following hypothesis by definition:

1. (There's nothing) -> (There's the absence of everything)
2. (There's nothing) -> (There's the absence of everything) -> (There's the absence of the absence of everything)¹ -> (There's everything) -> ~(There's the absence of everything)
And consider the following hypothesis by logic:
3. (There's nothing) v ~(There's nothing)²
4. ~[(There's the absence of everything) ^ ~(There's the absence of everything)]³
By 1 and 2, we have:
5. (There's nothing) -> (There's the absence of everything) ^ ~(There's the absence of everything)
By 5 and 3, we have:
6. [(There's the absence of everything) ^ ~(There's the absence of everything)] v ~(There's nothing)
By 6 and 4, we have:
7. ~(There's nothing)
Q.E.D.

¹ - Cause "everything" includes the "absence of everything", since "absence of everything" is something.
² - Law of excluded middle
³ - Law of non-contradiction

As a result, the space is not full of nothing. Cause nothing does not exist.

Is the demonstration right?
 
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Was this a joke? Because all you have is one long play on words.
 
No, it's not a joke.

Long play on words you say, so let me take off the words:

Let A and B be propositions.

Proving ~A

Consider the following hypothesis by definition:

1. A -> B
2. A -> ~B
And consider the following hypothesis by logic:
3. (A v ~A)²
4. ~(B ^ ~B)³
By 1 and 2, we have:
5. A -> (B ^ ~B)
By 5 and 3, we have:
6. (B ^ ~B) v ~A
By 6 and 4, we have:
7. ~A
Q.E.D.

² - Law of excluded middle
³ - Law of non-contradiction
 
charlie_sheep said:
No, it's not a joke.

Long play on words you say, so let me take off the words:

Let A and B be propositions.

Proving ~A

Consider the following hypothesis by definition:

1. A -> B
2. A -> ~B

From these two lines it follows that B and ~B, giving you a contradiction. Given a contradiction, you can derive any conclusion.
 
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