**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 *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 *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 *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, , a formula about natural numbers, corresponds to "there is no proof of *F*(*G*(*F*))".

We now define the formula *P*(*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*))= corresponds to "there is no proof of *F*(*G*(*F*))", as we have seen.

Consider the formula *P*(*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 *P*(*G*(*P*), is provable.

Suppose *P*(*G*(*P*))= is provable. Let *n* be the Gödel number of a proof of *P*(*G*(*P*)). Then, as seen earlier, the formula *q*(*n*,*G*(*P*)) is provable. Proving both

*q*(*n*,*G*(*P*)) and violates the consistency of the formal theory. We therefore conclude that *P*(*G*(*P*) is not provable.

Consider any number *n*. Suppose *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 *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*)), *P*(*G*(*P*)) = , is provable. Proving both , and *q*(*n*,*G*(*P*)), for all natural numbers *n*, violates ω-consistency of the formal theory. Thus if the theory is ω-consistent, *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

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.