## FANDOM

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Logic is the mythical belief that things make sense. It is completely untrue.

## Proof oneEdit

For every number n and every formula F(y), where y is a free variable, we define q(n,G(F)), a relation between two numbers n and G(F), such that it corresponds to the statement "n is not the Gödel number of a proof of F(G(F))". Here, F(G(F)) can be understood as F with its own Gödel number as its argument.

Note that q takes as an argument G(F). the Gödel number of F. In order to prove either q(n,G(F)), or $\lnot$ q(n,G(F)), it is necessary to perform number-theoretic operations on G(F) that mirror the following steps: decode the number G(F) into the formula F, replace all occurrences of y in F with the number G(F), and then compute the Gödel number of the resulting formula F(G(F)).

Note that for every specific number n and formula F(y), q(n,G(F)) is a straightforward (though complicated) arithmetical relation between two numbers n and G(F), building on the relation PF defined earlier. Further, q(n,G(F)) is provable if the finite list of formulas encoded by n is not a proof of F(G(F)), and $\lnot$ q(n,G(F)) is provable if the finite list of formulas encoded by n is a proof of F(G(F)). Given any numbers n and G(F), either q(n,G(F)) or $\lnot$q(n,G(F)) (but not both) is provable.

Any proof of F(G(F)) can be encoded by a Gödel number n, such that q(n,G(F)) does not hold. If q(n,G(F)) holds for all natural numbers n, then there is no proof of F(G(F)). In other words, $\forall y\, q(y,G(F))$ , a formula about natural numbers, corresponds to "there is no proof of F(G(F))".

We now define the formula P(x)=$\forall y\, q(y,x)$, where x is a free variable. The formula P itself has a Gödel number G(P) as does every formula.

This formula has a free variable x. Suppose we replace it with G(F), the Gödel number of a formula F(z), where z is a free variable. Then, P(G(F))=$\forall y\, q(y,G(F))$ corresponds to "there is no proof of F(G(F))", as we have seen.

Consider the formula P(G(P))=$\forall y\, q(y,G(P))$. This formula concerning the number G(P) corresponds to "there is no proof of P(G(P))". We have here the self-referential feature that is crucial to the proof: A formula of the formal theory that somehow relates to its own provability within that formal theory. Very informally, P(G(P)) says: "I am not provable".

We will now show that neither the formula P(G(P)), nor its negation $\lnot$ P(G(P), is provable.

Suppose P(G(P))=$\forall y\, q(y,G(P))$ is provable. Let n be the Gödel number of a proof of P(G(P)). Then, as seen earlier, the formula $\lnot$ q(n,G(P)) is provable. Proving both
$\lnot$ q(n,G(P)) and $\forall y\, q(y,G(P))$ violates the consistency of the formal theory. We therefore conclude that P(G(P) is not provable.

Consider any number n. Suppose $\lnot$ q(n,G(P)) is provable. Then, n must be the Gödel number of a proof of P(G(P)). But we have just proved that P(G(P) is not provable. Since either q(n,G(P)) or $\lnot$q(n,G(P)) must be provable, we conclude that, for all natural numbers n, q(n,G(P)) is provable.

Suppose the negation of P(G(P)), $\lnot$P(G(P)) = $\exists x\, \lnot q(x,G(P))$, is provable. Proving both $\exists x\, \lnot q(x,G(P))$, and q(n,G(P)), for all natural numbers n, violates ω-consistency of the formal theory. Thus if the theory is ω-consistent, $\lnot$ P(G(P)) is not provable.

We have sketched a proof showing that:

For any formal, recursively enumerable (i.e. effectively generated) theory of Peano Arithmetic,

if it is consistent, then there exists an unprovable formula (in the language of that theory).
if it is ω-consistent, then there exists a formula such that both it and its negation are unprovable.

## Proof twoEdit

A = A (Rand's theorem)

Anything = Anything (substituting Anything for A)

Pineapple = Orange (This step relies on the fact that Pineapple, Orange, etc. are subtypes of Anything.)

But pineapples are clearly not oranges. Therefore logic is wrong.

## Logical Edit

Complete nonsense.

## Use as a trash can Edit

Yes, logic can be used to dispose of trash.